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Among all rectangles that have a perimeter of 156, find the dimensions of the one whose area is largest. write your answers as fractions reduced to lowest terms.

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mathstudent55
P = 2L + 2W

We have a perimeter of 156, so we have

2L + 2W = 156

Let the length = x

2x + 2W = 156

The width is

2W = 156 - 2x

W = 78 - x

The area of a rectangle is A = LW

A = x(78 - x)

A = 78x - x^2

This is an inverted parabola, so there is a maximum value.

78x - x^2 = 0

x(78 - x) = 0

x = 0 or x = 78

The zeros of the parabola are at x = 0 and x = 78.
Since the parabola is symmetric over its vertical axis, the maximum values occurs at the x-value in the middle of 0 to 78, which is 39.

At x = 39, the area has a maximum value.

L = 39 & W = 39

It's a square with side measuring 39.
abidemiokin

The dimensions of the one whose area is largest is 39 by 39

The formula for calculating the perimeter of a rectangle is expressed as:

P = 2(L + W)

L is the length of the rectangle

W is the width

Given that the perimeter is 156, hence;

2(L + W) = 156

2L + 2W = 156

L + W = 78

W = 78 - L

The area of the rectangle is expressed as:

A = LW

A = L(78-L)

A = 78L - L^2

To get the dimensions of the one whose area is largest, dA/dL = 0

dA/dL = 78 - 2L

0 = 78 - 2L

2L = 78

L = 78/2

L = 39

Recall that 2(L+W) = 156

2(39 + W) = 156

39 + W = 78

W = 78 - 39

W = 39

Hence the dimensions of the one whose area is largest is 39 by 39

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