Question

A man with 200 ft. of fencing material wishes to fence off an area in the shape of a rectangle. What should be the dimensions of the area if the enclosed space is to be as large as possible? What is the largest area? Hint: A = lw, P = 2l + 2w

Answer 1

Step-by-step explanation:

200 ft. is the perimeter, since that's how much fencing there is. 2 lengths and 2 widths of the field = 200: 2L+2w=200.

We also know the area of a rectangle is length times width, and we want to maximize that. A=L(w).

We can't solve a single equation with two variables, so we can eliminate either one in the first equation. For no particular reason, let's solve for L:

2L+2w=200. Subtract 2w from both sides: 2L=200-2w.

Divide through by 2 to get L:

L=100-w.

Substitute (100-w) for L in the area equation: A=(100-w)w.

Distribute the w, rearrange the terms, and A= -w2 + 100w

Graph this to find your vertex. (You can also use -b/2a=x to find the x coordinate of your vertex then plug it in to get the y.)

The maximum point is at (50, 2500), which is (w, A) in variables. So the width at maximum area 2500 ft2 is 50 ft.

Squares always maximize area! The field should be 50 ft x50 ft, and the area is 2500 ft2.

Answer 2

Given :  A man with 200 ft. of fencing material wishes to fence off an area in the shape of a rectangle.

Hint: A = lw, P = 2l + 2w

To Find : What should be the dimensions of the area if the enclosed space is to be as large as possible?

What is the largest area?

Solution:

A = lw

A = area  in sq ft

l = length   in ft

w = width  in ft

P = 2l + 2w   in ft

P = perimeter = 200 ft

=> 2l + 2w  = 200

=> l + w = 100

=> l = 100 - w

A = lw   = (100 - w)w

=> A = 100w  -  w²

=> dA/dw  = 100  - 2w

dA/dw = 0

=> 100 - 2w = 0

=> w = 50

d²A/dw²  = - 2 < 0

Hence A is maximum when w = 50

l = 100 - w = 100 - 50 = 50

Hence Dimensions of the area = 50 ft by 50 ft

Area = 50 * 50  = 2500 sq ft

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