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Prove that the most efficient way to build a rectangular fenced area along a river-so that only three sides of fencing are needed-is to make the side parallel to the river twice as long as the other sides. You may assume that you have a fixed amount of fencing material.

Short Answer

Expert verified

The area is maximized when x=P4, it implies that when the rectangle is a square.

Step by step solution

01

Given information.

The goal is to demonstrate that, given a specified perimeter P, the rectangle with the largest feasible area is always a square.

02

Explanation of the statement.

Consider, xand y are the sides of the rectangle.

The rectangle's perimeter is then P=2x+2y.

So,

y=12(P-2x)

The area is as follows.

A=xy=x12(P-2x)

The area's derivation is as follows:

A=x12(P-2x)A'=12P-2x=0x=P4

The only critical point is x=P4.

since,A'P4-1>0and A'P4+1<0.

The first derivative so indicates that A(x)consists of a local maximum at x=P4.

The highest possible value is at 0,P2.

The area is therefore maximised when x=P4, i.e if the rectangle is square in shape.

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Most popular questions from this chapter

In Exercises 37โ€“42, find v and find the unit vector in the direction of v.

v=13,-23

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(b) True or False: โˆ‘k=0nโ€Š1k+1+โˆ‘k=1nโ€Šk2is equal to โˆ‘k=0nโ€Šk3+k2+1k+1.

(c) True or False: โˆ‘k=1nโ€Š1k+1+โˆ‘k=0nโ€Šk2is equal to โˆ‘k=1nโ€Šk3+k2+1k+1 .

(d) True or False: โˆ‘k=1nโ€Š1k+1โˆ‘k=1nโ€Šk2 is equal to โˆ‘k=1nโ€Šk2k+1.

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