saneayahsimmons
16.07.2019 •
Mathematics
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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Ответ:
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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Ответ:
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.
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