Applied Optimization: Videos & Practice Problems

In applied optimization problems, we often seek to maximize or minimize a specific value, such as area, under given constraints. A classic example involves using a fixed amount of fencing to create a rectangular area, where one side is formed by an existing structure, like a rock wall. In this scenario, we have 200 feet of fencing available to construct three sides of a rectangle, and our goal is to determine the dimensions that yield the maximum area.

To begin, we can express the area \( A \) of the rectangle as a function of its dimensions. The area is calculated as the product of length and width, represented by the equation \( A = x \cdot y \), where \( x \) is the length and \( y \) is the width. Given the constraint of 200 feet of fencing, we can derive a relationship between \( x \) and \( y \) based on the perimeter. The perimeter for our three sides can be expressed as \( 2x + y = 200 \). By rearranging this equation, we find \( y = 200 - 2x \). Substituting this expression for \( y \) into the area function gives us \( A = x(200 - 2x) = 200x - 2x^2 \).

Next, we need to determine the domain of our function. Since dimensions cannot be negative, we establish that \( x \) must be greater than or equal to 0. Additionally, since the total fencing is 200 feet, if both lengths \( x \) are maximized to 100 feet, then \( y \) would be 0, leading to the restriction \( x \leq 100 \). Thus, our domain is \( 0 \leq x \leq 100 \).

To find the maximum area, we calculate the critical points by taking the derivative of the area function. The first derivative is \( A' = 200 - 4x \). Setting this equal to zero to find critical points, we solve \( 200 - 4x = 0 \), yielding \( x = 50 \). We must also evaluate the endpoints of our domain, \( x = 0 \) and \( x = 100 \).

Now, we evaluate the area at these points: - For \( x = 0 \): \( A(0) = 200(0) - 2(0)^2 = 0 \)- For \( x = 100 \): \( A(100) = 200(100) - 2(100)^2 = 0 \)- For \( x = 50 \): \( A(50) = 200(50) - 2(50)^2 = 5000 \)The maximum area occurs at \( x = 50 \), resulting in an area of 5000 square feet. To find the corresponding width \( y \), we substitute \( x = 50 \) back into the equation for \( y \): \( y = 200 - 2(50) = 100 \). Therefore, the dimensions that maximize the area are 50 feet by 100 feet.

In summary, when solving applied optimization problems, it is essential to define the function to be optimized, establish constraints, determine the domain, find critical points, and evaluate the function at these points to identify maximum or minimum values. Depending on whether the interval is closed or open, the method for determining these values may vary, utilizing either the extreme value theorem or the second derivative test.

In applied optimization problems, one common task is to minimize the surface area of a three-dimensional object. For instance, consider a shipping company that needs to construct a rectangular box with a square base, designed to hold a volume of 125 cubic inches without a lid. The goal is to determine the dimensions that will minimize the amount of cardboard used, which directly correlates to minimizing the surface area of the box.

To begin, we define the variables: let x represent the length of each side of the square base, and y denote the height of the box. The surface area S of the box can be expressed as:

$$ S = 4xy + x^2 $$

Here, 4xy accounts for the four sides of the box, and x² represents the area of the base. Given the constraint of a fixed volume, we know:

$$ V = x^2y = 125 $$

From this equation, we can solve for y:

$$ y = \frac{125}{x^2} $$

Substituting this expression for y back into the surface area equation gives us:

$$ S = 4x\left(\frac{125}{x^2}\right) + x^2 = \frac{500}{x} + x^2 $$

Next, we need to find the critical points of this function to determine where the surface area is minimized. We start by finding the derivative of S:

$$ S' = -\frac{500}{x^2} + 2x $$

Setting the derivative equal to zero to find critical points:

$$ -\frac{500}{x^2} + 2x = 0 $$

Rearranging gives:

$$ \frac{500}{x^2} = 2x $$

Multiplying both sides by x² leads to:

$$ 500 = 2x^3 $$

Dividing by 2 results in:

$$ x^3 = 250 $$

Taking the cube root yields:

$$ x = \sqrt[3]{250} \approx 6.3 $$

To confirm that this critical point is indeed a minimum, we can analyze the second derivative:

$$ S'' = \frac{1000}{x^3} + 2 $$

Since both terms are positive for x > 0, the second derivative is positive, indicating that the critical point corresponds to a minimum surface area.

Now, substituting x back into the equation for y gives:

$$ y = \frac{125}{(6.3)^2} \approx 3.15 $$

Thus, the dimensions of the box that minimize the surface area are approximately 6.3 inches for each side of the base and 3.15 inches for the height. This approach illustrates the process of applying optimization techniques to real-world problems, reinforcing the importance of understanding both the mathematical principles and their practical applications.

In applied optimization problems, particularly in maximizing revenue or profit for a business, the process involves expressing the desired value as a function and then identifying its maximum or minimum. To illustrate this, consider a coffee shop that sells x coffees per day at a price of p dollars each. The price demand function is given by:

$P\left(x\right)=100-x4$

To maximize daily revenue, we first need to express revenue as a function of x. Revenue (R) is calculated as the product of the price per coffee and the number of coffees sold:

$R\left(x\right)=P\left(x\right)x$

Substituting the price function into the revenue function gives:

Distributing x results in:

$R\left(x\right)=100x-x{2}^2$

Next, we need to determine the domain restrictions for x. Since x represents the number of coffees sold, it must be non-negative:

$x>=0$

Additionally, the price function must also be non-negative, leading to the inequality:

$100-\backslash frac\{x\}\{4\}\; \backslash geq\; 0$

Solving this inequality results in:

$x<=400$

Thus, the domain for x is:

With the revenue function established and the domain restrictions identified, the next steps involve finding critical points and applying either the extreme value theorem or the second derivative test to determine the maximum revenue. This process is essential for businesses aiming to optimize their sales strategies effectively.

In this analysis of revenue maximization, we begin with the revenue function defined as the product of price and quantity sold, expressed mathematically as:

$$ R(x) = 100x - \frac{x^2}{4} $$

Here, \( x \) represents the number of items sold, with the domain restrictions set to \( 0 \leq x \leq 400 \). This ensures that both the quantity sold and the price remain non-negative.

To find the critical points, we first compute the first derivative of the revenue function:

$$ R'(x) = 100 - \frac{x}{2} $$

Setting the first derivative equal to zero allows us to identify critical points:

$$ 100 - \frac{x}{2} = 0 $$

Solving for \( x \) gives:

$$ x = 200 $$

Next, we evaluate the revenue at the endpoints of the domain and at the critical point to determine the maximum revenue. We calculate:

1. \( R(0) = 100(0) - \frac{0^2}{4} = 0 \)

2. \( R(200) = 100(200) - \frac{200^2}{4} = 10,000 \)

3. \( R(400) = 100(400) - \frac{400^2}{4} = 0 \)

From these calculations, we observe that the maximum revenue occurs at the critical point \( x = 200 \), yielding a maximum revenue of:

$$ R(200) = 10,000 $$

This indicates that to achieve the maximum revenue of $10,000, 200 items must be sold.

Alternatively, one could apply the second derivative test to confirm the nature of the critical point. The second derivative is calculated as:

$$ R''(x) = -\frac{1}{2} $$

Since the second derivative is negative, it confirms that the critical point at \( x = 200 \) is indeed a maximum. This method provides an efficient way to verify the results obtained from evaluating the endpoints and critical points.

In summary, the analysis reveals that selling 200 items maximizes revenue at $10,000, demonstrating the importance of critical point analysis in optimization problems.

In this scenario, we explore how to maximize profit for a business selling custom-made phone cases. The price per case, denoted as \( p(x) \), is influenced by the number of cases sold, represented by the function \( p(x) = 50 - 0.2x \), where \( x \) is the number of cases sold in dollars. The total cost to produce \( x \) cases is given by the cost function \( C(x) = 100 + 15x \).

To find the maximum profit, we first need to establish the profit function, \( P(x) \), which is defined as revenue minus cost. The revenue \( R(x) \) can be calculated as the product of price and quantity sold, leading to the equation:

\[ R(x) = p(x) \cdot x = (50 - 0.2x) \cdot x = 50x - 0.2x^2 \]

Thus, the profit function becomes:

\[ P(x) = R(x) - C(x) = (50x - 0.2x^2) - (100 + 15x) \]

After simplifying, we arrive at:

\[ P(x) = -0.2x^2 + 35x - 100 \]

Next, we determine the domain restrictions for \( x \). Since selling a negative number of cases is not feasible, we establish that \( x \geq 0 \). Additionally, to ensure the price remains non-negative, we solve the inequality \( p(x) \geq 0 \), leading to \( 50 - 0.2x \geq 0 \) or \( x \leq 250 \). Therefore, the domain for \( x \) is \( 0 \leq x \leq 250 \).

To find the critical points, we calculate the first derivative of the profit function:

\[ P'(x) = -0.4x + 35 \]

Setting the first derivative equal to zero to find critical points gives:

\[ -0.4x + 35 = 0 \implies x = \frac{35}{0.4} = 87.5 \]

Since we cannot sell half a phone case, we round this to the nearest whole number, resulting in \( x = 88 \).

To confirm that this point yields a maximum profit, we compute the second derivative:

\[ P''(x) = -0.4 \]

Since the second derivative is negative, it indicates that the profit function is concave down, confirming that \( x = 88 \) is indeed a maximum.

Now, we can calculate the maximum profit by substituting \( x = 88 \) back into the profit function:

\[ P(88) = -0.2(88^2) + 35(88) - 100 = 1431.20 \]

Thus, the maximum profit is $1,431.20. To find the corresponding price at this maximum profit, we substitute \( x = 88 \) into the price function:

\[ p(88) = 50 - 0.2(88) = 32.40 \]

Therefore, the optimal price to charge per phone case for maximizing profit is $32.40. In summary, to achieve maximum profit, Gerald and Katya should sell 88 phone cases at a price of $32.40, resulting in a profit of $1,431.20.

In the context of inventory management, businesses often face the challenge of minimizing costs associated with storing and reordering products. This scenario is typically framed as an inventory control problem, where the goal is to determine the optimal order quantity that minimizes total costs. To illustrate this, consider a math store that sells 400 calculators annually, incurs a storage cost of $2 per calculator per year, and has a fixed reorder cost of $25.

To solve this problem, we first define a variable, x, representing the number of calculators ordered each time. The total cost function comprises two components: the storage cost and the reorder cost. The storage cost can be calculated as the product of the storage cost per item and the average number of items in storage. Since the average number of items in storage is x/2, the storage cost becomes:

Storage Cost = $2 * (x/2) = $x

Next, we calculate the reorder cost, which is determined by the number of orders placed throughout the year. The number of orders is the total annual sales divided by the number of items per order, or 400/x. Thus, the reorder cost is:

Reorder Cost = $25 * (400/x) = $10,000/x

Combining these two components, the total cost function can be expressed as:

Total Cost (C) = x + 10,000/x

To ensure the solution is valid, we must establish domain restrictions for x. Since the store cannot order a negative number of calculators or zero calculators, x must be at least 1. Additionally, as the store sells 400 calculators per year, x cannot exceed 400. Therefore, the domain for x is:

1 ≤ x ≤ 400

With the cost function and domain established, the next steps involve finding critical points through differentiation and applying optimization techniques such as the first derivative test or the second derivative test. It is important to note that the final value of x should be a whole number, as ordering a fraction of a calculator is not feasible. If the calculated value of x is a decimal, the nearest whole numbers should be tested to determine the optimal order quantity.

By following these steps, businesses can effectively minimize their inventory costs and enhance their operational efficiency.

To minimize costs in inventory management, we start by establishing a cost function and identifying domain restrictions. In this scenario, the cost function is defined, and we need to find critical points by taking the derivative. The derivative of the cost function, denoted as \( c'(x) \), is calculated as:

\[ c'(x) = 1 - \frac{10,000}{x^2} \]

Setting the derivative equal to zero allows us to find critical points:

\[ 1 = \frac{10,000}{x^2} \]

By rearranging and solving for \( x^2 \), we find:

\[ x^2 = 10,000 \]

Taking the square root gives us:

\[ x = \pm 100 \]

However, considering the domain restriction that \( x \) must be greater than or equal to 400, we discard the negative value and focus on \( x = 100 \) as our critical point.

Next, we apply the Extreme Value Theorem since we are working within a closed interval defined by the endpoints 400 and the critical point 100. We evaluate the cost function at these points to determine the minimum cost:

Calculating the cost function at the endpoints and the critical point:

\[ c(1) = 1 + \frac{10,000}{1} = 10,001 \]

\[ c(100) = 100 + \frac{10,000}{100} = 200 \]

\[ c(400) = 400 + \frac{10,000}{400} = 425 \]

Among these values, the minimum cost is \( 200 \), which occurs when ordering 100 calculators per order. This means that if the company orders 100 calculators each time, they will incur the lowest cost.

To determine how many orders the company should place annually, we divide the total demand of 400 calculators by the order size of 100:

\[ \text{Number of Orders} = \frac{400}{100} = 4 \]

Thus, the company should place 4 orders per year to optimize their inventory costs effectively. This analysis highlights the importance of balancing order size and frequency to minimize overall expenses in inventory management.

The Norman window is a unique shape consisting of a rectangle topped with a semicircle. In optimization problems involving this window, the goal is to maximize the amount of light that passes through, which can be approached by maximizing the area of the window. Given that the perimeter of the window is 24 meters and the semicircular part allows only one-quarter of the light per square meter compared to the rectangular part, we can derive an equation to represent the area of the window.

To begin, we define the width of the rectangle as \( x \) and the height as \( y \). The radius of the semicircle is \( \frac{x}{2} \). The total area \( A \) of the window can be expressed as the sum of the area of the rectangle and one-quarter of the area of the semicircle. The area of the rectangle is given by \( A_{rectangle} = x \cdot y \), while the area of the semicircle is \( A_{semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{x}{2}\right)^2 = \frac{\pi x^2}{8} \). Therefore, the area function becomes:

\[A = xy + \frac{1}{4} \cdot \frac{\pi x^2}{8} = xy + \frac{\pi x^2}{32}\]

Next, we need to express \( y \) in terms of \( x \) using the perimeter constraint. The perimeter \( P \) of the window is given by:

\[P = x + 2y + \frac{\pi x}{2} = 24\]

Rearranging this equation to solve for \( y \) yields:

\[2y = 24 - x - \frac{\pi x}{2} \implies y = 12 - \frac{x}{2} - \frac{\pi x}{4}\]

Substituting this expression for \( y \) back into the area function gives us:

\[A = x\left(12 - \frac{x}{2} - \frac{\pi x}{4}\right) + \frac{\pi x^2}{32}\]

Distributing \( x \) results in:

\[A = 12x - \frac{x^2}{2} - \frac{\pi x^2}{4} + \frac{\pi x^2}{32}\]

Combining like terms leads to a simplified area function:

\[A = 12x - x^2\left(\frac{1}{2} + \frac{\pi}{4} - \frac{\pi}{32}\right)\]

Next, we determine the domain for \( x \). Since \( x \) must be non-negative and cannot exceed the perimeter constraints, we find that \( 0 \leq x \leq 9.34 \) meters. To find the critical points, we take the derivative of the area function and set it to zero:

\[A' = 12 - x\left(1 + \frac{\pi}{2} - \frac{\pi}{16}\right) = 0\]

Solving for \( x \) gives us a critical point of approximately \( x \approx 5.05 \) meters. Evaluating the area function at the endpoints \( x = 0 \) and \( x = 9.34 \), as well as at the critical point, allows us to identify the maximum area. The calculations yield:

\[A(0) = 0, \quad A(5.05) \approx 30.32, \quad A(9.34) \approx 8.51\]

Thus, the maximum area occurs at \( x = 5.05 \) meters. To find the corresponding height \( y \), we substitute \( x \) back into the equation for \( y \):

\[y = 12 - \frac{5.05}{2} - \frac{\pi \cdot 5.05}{4} \approx 5.51 \text{ meters}\]

In conclusion, to maximize the light passing through the Norman window, the optimal dimensions are approximately \( x = 5.05 \) meters for the width and \( y = 5.51 \) meters for the height.

In this applied optimization problem, we focus on designing a box with an open top by cutting equal-sized squares from the corners of a rectangular sheet of cardboard measuring 3 feet by 8 feet. The objective is to maximize the volume of the box formed by folding up the sides after the corners are cut. To achieve this, we need to derive a function that represents the volume based on the dimensions of the squares cut out, denoted as x.

First, we establish the dimensions of the box. The length of the box after cutting out the squares will be 8 - 2x (subtracting the squares from both ends of the 8-foot side) and the width will be 3 - 2x (subtracting from the 3-foot side). The height of the box, when the sides are folded up, is equal to x. Therefore, the volume V of the box can be expressed as:

$$ V = (8 - 2x)(3 - 2x)(x) $$

Expanding this expression, we find:

$$ V = x(24 - 22x + 4x^2) = 4x^3 - 22x^2 + 24x $$

Next, we determine the domain restrictions for x. Since the dimensions must be non-negative, we have x ≥ 0. Additionally, the maximum value for x is limited by the shorter side of the cardboard. Since the maximum square cut from the 3-foot side is 1.5 feet, we conclude that x ≤ 1.5. Thus, the domain for x is [0, 1.5].

To find the critical points, we take the derivative of the volume function:

$$ V' = 12x^2 - 44x + 24 $$

Factoring out a 4 gives:

$$ V' = 4(3x^2 - 11x + 6) $$

Setting the derivative equal to zero, we apply the quadratic formula to solve for x:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Substituting a = 3, b = -11, and c = 6 results in:

$$ x = \frac{11 \pm \sqrt{(-11)^2 - 4(3)(6)}}{2(3)} = \frac{11 \pm 7}{6} $$

This yields two critical points: x = 3 and x = \frac{2}{3}. However, only x = \frac{2}{3} \approx 0.67 is within the valid domain.

To find the maximum volume, we evaluate the volume function at the critical point and the endpoints of the interval:

1. At x = 0: V(0) = 0

2. At x = 1.5: V(1.5) = 0

3. At x = \frac{2}{3}: Plugging this value into the volume function gives:

$$ V\left(\frac{2}{3}\right) = 4\left(\frac{2}{3}\right)^3 - 22\left(\frac{2}{3}\right)^2 + 24\left(\frac{2}{3}\right) $$

Calculating this results in approximately 7.41 cubic feet.

Thus, the maximum volume of the box is 7.41 cubic feet, achieved when the dimensions of the squares cut from the corners are 0.67 feet. This solution highlights the importance of understanding the relationships between dimensions and volume in optimization problems.

In this optimization problem, we aim to determine the shortest length of rope that can be tied from the ground, over the top of a tree, to a cliff. We have a 10-meter tall tree positioned 5 meters away from the base of a rock cliff. To solve this, we first establish a diagram and identify our variables: let \( l \) represent the length of the rope, \( h \) the height of the cliff, and \( x \) the horizontal distance from the tree to the point directly below the top of the cliff.

To express \( l \) in terms of \( x \) and \( h \), we recognize that the length of the rope forms a right triangle where \( l^2 = h^2 + (x + 5)^2 \). Taking the square root gives us \( l = \sqrt{h^2 + (x + 5)^2} \). To minimize \( l \), we need to express \( h \) in terms of \( x \). Using the similarity of triangles, we derive the relationship \( \frac{h}{10} = \frac{x + 5}{x} \), leading to \( h = \frac{10(x + 5)}{x} \). Simplifying this gives \( h = 10 + \frac{50}{x} \).

Substituting \( h \) back into the equation for \( l \) results in \( l = \sqrt{\left(10 + \frac{50}{x}\right)^2 + (x + 5)^2} \). Next, we find the critical points by taking the derivative of \( l \) and setting it to zero. The derivative, using the chain rule, is complex but can be simplified to find where the numerator equals zero. This leads us to the equation \( x^3 - 500 = 0 \), giving us \( x = \sqrt[3]{500} \approx 7.98 \).

Since \( x \) must be greater than zero, we discard any negative solutions. We then substitute \( x = \sqrt[3]{500} \) back into the equation for \( l \) to find the minimum length of the rope. This results in \( l \approx 20.81 \) meters. Thus, the shortest length of rope that can be tied from the ground over the top of the tree to the cliff is approximately 20.81 meters.