10. Catching rainwater A 1125 ft^3 open-top rectangular tank w...

Question

10. Catching rainwater A 1125 ft^3 open-top rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.
a. If the total cost is c=5(x^2+4xy) + 10xy, what values of x and y will minimize it?
b. Give a possible scenario for the cost function in part (a).

a. If the total cost is c=5(x^2+4xy) + 10xy, what values of x and y will minimize it?

b. Give a possible scenario for the cost function in part (a).

Accepted Answer

Hello. In this video, we are going to be approaching the following optimization problem. So we are told that a 576 inch cubed open top box storage unit with a base of X inches on a side and Y inches deep is to be constructed for an industrial warehouse. The costs include both the material for the base and the four side walls, as well as handling charges proportional to the product X and Y. If the total cost is C equal to 6 multiplied by X2 + 4 XY plus 12 XY, what values of X and Y will minimize this cost? So, let's just go ahead and start by drawing a picture. Now, we are told that the shape we are dealing with is an open top box. So we have this box right here. And I apologize cause my drawing is not the best. And this is a box with an open top. So we have a box of 5 sides. Now we are told that this box has a base of X inches and the height of Y inches. Now because we're dealing with the box, we can assume that the base is going to be a square, and since the base is a square, that means all the sides of the base are equivalent to each other and will be labeled with X. Now, we don't know what the height of the box is, but we will label the height Y. So, what is going to be the volume of this box? Well, we know that the volume of a cube in general, is the volume is equal to length times width times height. We know that the length and width in this case are going to be the same, they're going to be X, and the height is going to be Y. Furthermore, the problem tells us that this volume is going to be 576 inches cubed. So by plugging in these values, we can write the volume equation to be X2 multiplied by Y is equal to 576. We will go ahead and set this equation to the side for now. Now, we are told that the total cost of this cube is given by the following equation. C is equal to 6 multiplied by X2 + 4 XY + 12 XY. Let's just go ahead and first simplify this cost function. We will start by distributing 6 into the quantity, and that will give us C as equal to 6 X 2 plus 24 XY plus 12 XY. Notice that we can combine the last two terms together, and that will give us a simplified cost formula to be C is equal to 6 X2 + 36 XY. Now, this is going to be the formula that we want to optimize. Specifically, we are trying to find the dimensions of X and Y that cause the function to be a minimum value. Now, what we are going to do is we can go ahead and simplify this function to be in terms of one variable. Notice that we have the last term to be X multiplied by Y. We can use our volume equation to create a substitution for Y in terms of X. So, if we solve for Y in our volume equation, we get Y is equal to 576 divided by X squared. This is going to be a substitution for Y that we can now plug into our cost function. So that is going to give us C is equal to 6 X squared plus 36X multiplied by the quantity. 576 divided by X squared. And if we multiply these two terms together, we can further simplify the cost function to be 6 X2 plus 20,736 divided by X. This is going to be the simplest form of the cost function, especially since we rewrote it in terms of X. Now what we want to do is we want to find the value of X that minimizes the cost. In order to do this, we are going to take the derivative with respect to X of the given cost function. By the power rule and by the quotient rule, we will have DC divided by DX equal to 12 X minus 20,736 divided by X squared. By combining these two terms together, we can further rewrite this derivative to be 12 x cubed minus 20,736 divided by X squared. Now, since we are trying to determine the value of X, that is a minimum value, we want to find the critical point of this function. In order to find the critical point, we will find the value of X that causes the function to be zero. So we will take the numerator. 12 x cubed minus 20,736 and set it equal to 0. By adding 20,736 then dividing by 12, we will end up with X cubed is equal to 1,728. And by taking the cube root of both sides of this equation, we get X is equal to 12 inches. Now what we need to do is we need to verify that this is a minimum value. In order to do this, we will use the first derivative test. We will take our critical 0.12 place it on a number line, and analyze the behaviors around 12. Testing points that we can plug into the derivative that are around 12 are going to be 11 and 13. When we plug an 11 into the derivative, we get a negative value. And when we plug in the value of 13 into the derivative, we get a positive value. As we pass through 12, the function shifts from decreasing to increasing, and this verifies that this value of X is going to be a minimum value. Now that we verify that X is equal to 12 inches is a minimum value, if we take this value of X and plug it into our substitution for Y, we can also find a value of Y that will also be a minimum value. That is going to give us Y is equal to 576 divided by 12 squared, and this is going to simplify to be 4 inches. So, to summarize, by substituting Y in terms of X into the cost, by taking the derivative and verifying that the critical point is a minimum value, we figured out that the dimensions of X and Y that cause the cost function to be a minimum value is going to be Y is equal to 4 inches and X is equal to 12 inches, and this is going to be the solution to the problem. So I hope this video helps you in understanding how to approach this problem, and I will go ahead and see you all in the next video.

Key Concepts

Optimization in Calculus

Partial Derivatives

Critical Points and Second Derivative Test

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