A man with 200ft. of fencing materials wishes to fence off an area in the shape of a rectangle. What should be the diemnsions of the area if the enclosed space is to be as large as possible?

Question

Additional question: what is the largest area?

There's a hint on it but i don't know to solve it

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

Answer 1

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. Have fun mathing, and message me if you have any questions!

Liz Z.

Answer 2

When maximizing the area, the idea is to make the shape a square. So since you have 200 ft of fencing then you need to divide it by 4 to see what is the length of each side of the square. 200/4=50 so each side will need to be 50 ft. Then to answer your next question, find the area of a square with side 50. I will let you finish the part b.

I hope this helps!

Mrs. S.