Question

Can any one solve this problem with solutions​

Answer 1

Answer:

Area is 2500 ft.2

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.

I hope this helps. :-)