Pen problems

b. A rancher plans to make four ...

Question

Pen problems

b. A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100 m² (see figure). What are the dimensions of each pen that minimize the amount of fence that must be used?

Accepted Answer

Welcome back, everyone. A farmer wants to create 3 identical and side by side rectangular vegetable patches against a fence, each with an area of 75 square meters. What are the dimensions of each patch that minimize the amount of fencing required? A says it's 2 by 37.5 m. C 25, 25 by 3 m, C 15 by 5 m, and the D 7.5 by 10 m. Now, let's try to picture our problem here to better understand what's going on. So we're saying that our farmer has 3 identical rectangular vegetable patches side by side, OK. And we know that each of these patches will have a length L and a width W, OK? So in that case, we have our width here. Each of these dimensions would be W and then the side of this, uh, this, this patch or the side of this configuration would also be L, OK? And now we know that he wants to minimize the amount of fencing that would be required based on the dimensions, OK? So how could we do that? Well, if we think about the total fencing that's needed here, then let's let L. Uh, sorry, let's let F. Be equal to the total fencing. OK. Then in that case F is going to be equal to 2 L plus 6 W. Now if we're going to minimize the amount of fencing, it means that we'll need to figure out some critical points and determine if they are maxima or minimal but what would be important here. Is that we need to express F in terms of one variable, OK? So how can we do that? Well, we're also told that each, each, uh, vegetable patch, OK, has an area of 75 square meters. In other words, the area here, OK, is 75 square meters. Now, we know that when it comes to area. The area of a rectangle is equal to the length multiplied by the width. So this would be LW, which equals 75 square meters, which then means that or sorry, L. Is going to be equal to 75 divided by W. So now if we substitute that into our formula for the total fencing, this means the total fencing F is going to be equal to 2 multiplied by 75 divided by W plus 6W, which is going to be equal to 150 divided by W. Plus 6 W. OK? So this is the expression for our fencing F. Now, here, since we have an expression for F in terms of W, we can, we can first maximize by setting the first derivative of F to zero and then solving for that value of W, OK? So setting. All right, let me see, minimizing. If right up properly here, mini. My F with respect to W. OK, then. The derivative of F with respect tow is going to be equal to the derivative of 150 divided by W. that's -150 divided by W squared plus the derivative of 6W, which is 6. I know that would be equal to 0. Now, in this case, the size for W, then this means 6 is gonna be equal to 150 divided by W squared. So W2 equals 150 divided by 6, which is 25, which means then that W is going to be equal to the square root of 25, which is 5. Now that we have our critical point, let's test if it is, let's test if it is a minimum value, if it's the minimum. And we can use that by using the second derivative test. So testing if WEA 5 corresponds to a minimum. Using the 2nd derivative test. OK. Then that means we can find the second derivative of F and if it is positive, that means the function is concave up confirming that the value is a minimum. So in this case, the second derivative for a total expression for the total fencing fencing would be equal to the derivative with respect to W of the first derivative -150 divided by W2 + 6. And this is gonna be equal to 300 divided by W cubed now. That means if we were to substitute 5 here when W equals 5, we can tell that our second derivative will be positive, OK, because our denominator will be 5 cubed and a positive divided by a positive will be positive. So in this case, the second derivative is 0, so that means this point corresponds or this value for WA equals 5 corresponds to a minimum. Now that we know it corresponds to a minimum and we know that it's the width of our fence, we can go back to south for the fence's length because no, that means then. That the length, which is 75 divided by W is gonna be 75 divided by 5, which equals 150. Therefore, OK. The dimensions of each vegetable patch that minimize the amount of fencing required are 15 m by 5 m. If we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Pen problems

b. A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100 m² (see figure). What are the dimensions of each pen that minimize the amount of fence that must be used? <IMAGE>

Key Concepts

Optimization

Area and Perimeter Formulas

Constraints

Watch next