Maximum area A line segment of length 10 joins the points (0, ...

Question

Maximum area A line segment of length 10 joins the points (0, p) and (q, 0) to form a triangle in the first quadrant. Find the values of p and q that maximize the area of the triangle.

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A right-angled triangle is formed in the first quadrant, with its right angle at the origin, one vertex along the Y axis. At 0. U in parentheses and another along the x axis at V.0 in parentheses. The hypotenuse of the triangle has a length of 12 units. Find the values of U and V that maximize the area of the triangle. Awesome. So it appears for this particular problem we're trying to figure out the values of U and V that maximize the area of the triangle, and that's what we're ultimately trying to solve for. We're trying to find these values of U and V, so two separate answers that maximize the area of the triangle. So with that said, our first step in order to help us solve this problem is let us create a diagram to help us better visualize the relationship between U, V, and the hypotenuse. So when we create a diagram, we should receive a graph that looks something along the line or not a graph, a diagram that looks something along the lines of this. As you can see, we have our point along the Y axis, which is our vertical axis, which is at 0.u in parentheses, and then on our X-axis, which is the horizontal. Lane, which is going to be V 0. So we could see, looking at our height of our triangle, which is one of our legs, will be in U units, because it follows the Y axis, whereas our X-axis, which is following along the horizontal axis, is going to be in V units. And then we're told that the hypotenuse of the triangle, which connects our two legs, which is denoted by a red line, is going to be 12 units. And let us note that we denoted the vertex on the Y axis in blue, whereas the vertex for the X on for the X-axis. Vertex for on the X-axis is going to be represented in green. Awesome. So now that we have a visualization of what's happening in this particular problem, let us note once again that the length of the hypotenuse of the right triangle is given by Pyres's theorem, so we can state that you squared. Plus V2 is going to be equal to 12 2, which means that we can say that U2 plus V2 is going to be equal to 144. Fantastic, that's what 12 squared is. So, thus we have the constraint that U2 plus V2 is equal to 144. So with that said, Our next step is we need to find the area of the triangle. So the area A of the right triangle is given by the following equation which states that A, the area is equal to 1/2, and we need to take 1/2, and we need to multiply it by the base and multiply it by the height. Which means that this would be equal to, for the case of our problem, 1/2 multiplied by V multiplied by U. Because V is our base and U is our height. So therefore, our area, without a doubt for this particular problem is going to be A is equal to 1/2 multiplied by U, multiplied by V. So now our next step is we need to maximize the area. So in order to maximize the area, as we should recall, we need to express the area as a function of one variable. So from the equation. U 2 plus V2 equals 144, we can solve for one variable in terms of the other. So let us solve for V in terms of U. So when we do that, we will find that V2d is equal to 144 minus U2. So isolating and solving for V all by itself will give us that V is equal to the square root of 144 minus u and once again you can use a little bit of algebra to do this rearranging to solve for V. Awesome. So V is equal to the square root of 144 minus U2. So now we need to substitute this expression for V into our area formula so we can state that a of you. is equal to 1/2 multiplied by you, multiplied by the square root of 144 minus U squared. So now at this point, we need to take the derivative of A of you. So in order to find the value of you that maximizes the area, we will, as we should recall, we will take the derivative of A of you with respect to you and set it equal to 0. So in order to do that, our first step is we need to apply the product and chain rule in order to perform this differentiation. So we will determine that a prime of you, where this prime symbol denotes the first derivative. So a of U is going to be equal to 1/2, and then we need to take 1/2 multiplied by the square root of 144 minus U squared. Plus you Multiplied by D by DU of the square root of 144 minus U squad. Fantastic. So now we need to differentiate the square root of 144 minus U2 using the chain rule. So when we do that, we will find that D by DU of the square root of 144 minus U2 is going to be equal to when we have it in its most simple form, negative U divided by the square root of. 144 minus U2. So thus, without a doubt, we could say that a of U is equal to 1/2 multiplied by the square root of 144 minus U2 minus U quad. Divided by 144. Minus U2. Fantastic. So moving right along here. Our next step now is we need to set the derivative equal to 0, so we need to set a U equal to 0 in order for us to find our critical points. So we need to take the square root of 144 minus U2. Minus U2 divided by the square root of 144 minus U2 is all equal to 0. So Now we need to isolate and solve for you all by itself. So I'm going to skip a few steps in order to do this because you should know the algebraic method to rearrange this isolate and solve for you, but as a hint, you need to multiply through by the square root of 144 minus U2, and then when you do that and you simplify, you will find that you all by itself is going to be equal to. 6 multiplied by the square root of 2, and that is our final answer for you. Hooray, we did it. But in order to make sure that ultimately this is the correct final answer before we can move on, we need to quickly verify this maximum. So to confirm that this critical point is a maximum, we need to check the second derivative of D squared. A by DU squared. Awesome. So with that said, we need to take, so we're gonna take D. Squared A by DU 2 is equal to, once again, we're taking the second derivative, so we can say that this is equal to U, multiplied by U2 minus 216, all divided by 144 minus U squared. All to the power of three halves. Fantastic. So now we need to note that since D 2 A by DU 2 is less than 0, for U is equal to 6 multiplied by the square rooted 2, fantastic. And because this is the case, we can say therefore, the function is going to be concave down, which I'll call a CD for short, and draw an arrow pointing down. So this function will be concave down, so this confirms that U equals 6 multiplied by the square rooted 2 is a maximum, so I'll give it a red check mark. So it is a maximum. So now we need to solve for the value of V. So now we're solving for V. So now we need to substitute in the value of U equals 6 multiplied by the square rooted 2 into our constraint equation. Which when we do that, we will find that 6 multiplied by the square rooted 2. Squared plus V2 equals 144, and using a little bit of algebra because we're now rearranging this equation to isolate and solve for V, you will find that V is going to be also equal to 6 multiplied by the square rooted 2. Hooray, we did it. That is our final answer. So in conclusion, we can state that the values of U and V that maximize the area of the triangle will be U equals 6 multiplied by the square root of 2, and V equals 6 multiplied by the square root of 2. So looking at our multiple choice answers, we will find Or rather, there's no multiple choice answers, so we're great. So let's quickly recap looking at the problem itself. So, once again, our final answers have to be, so let's wrap it up. So U is equal to 6 multiplied by the square rooted 2. And V is going to be equal to 6 multiplied by the square rooted 2, and those are our final two answers. We've officially solved for this problem. Hooray. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Maximum area A line segment of length 10 joins the points (0, p) and (q, 0) to form a triangle in the first quadrant. Find the values of p and q that maximize the area of the triangle.

Key Concepts

Area of a Triangle

Optimization

Constraints

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