1. The owner of the Rancho Los Feliz has 2640 yd of fencing material with which

Question

1. The owner of the Rancho Los Feliz has 2640 yd of fencing material with which to enclose a rectangular piece of grazing land along the straight portion of a river. If fencing is not required along the river, what are the dimensions of the largest area that he can enclose?

What is this area?
??? yd2

2. The owner of the Rancho Los Feliz has 3000 yd of fencing to enclose a rectangular piece of grazing land along the straight portion of a river and then subdivide it by means of a fence running parallel to the sides. No fencing is required along the river. (See the figure below.) What are the dimensions of the largest area that can be enclosed?
x = ???yd
y = ??? yd

What is this area?
???yd2

3. By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 16 in. long and 8 in. wide, find the dimensions of the box that will yield the maximum volume. (Round your answers to two decimal places.)

Explanation / Answer

Solution of first problem,

In the first problem we have given that Fensing 2640 yards.

We know that the area of the rectangle = A = length*Width = l * w

And the constraint = 2w + l = 2640

Solving for l = 2640 - 2w

Now plug this in the A = l*w = (2640 - 2w )*w

So A = 2640w -2w^2

Now we have to take derivative of A = 0 and solve for w.
So it becomes A' = 2640 - 4w = 0

4w = 2640 so    w = 660

length , l = (2640 - 2w) = 2640 - 2*660 = 1320

So we get l = 1320 yards and w = 660 yards

So the area of the rectangular piece of grazing land is = A = l*w = 1320 * 660 = 871200 yards^2

this is the required solution for first problem.

Solution of the second problem.

we have given that Fensing 3000 yards.

We know that the area of the rectangle = A = length*Width = l * w

And the constraint = 2w + l = 3000

Solving for l = 3000 - 2w

Now plug this in the A = l*w = (3000 - 2w )*w

So A = 3000w -2w^2

Now we have to take derivative of A = 0 and solve for w.
So it becomes A' = 3000 - 4w = 0

4w = 3000 so    w = 750

length , l = (3000 - 2w) = 2640 - 2*750 = 1500

So we get l = 1500 yards and w = 750 yards

So the area of the rectangular piece of grazing land is = A = l*w = 1500*750 = 1125000yards^2

This is the required answer for the second problem.

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