Find the length and width of a rectangle that has the given perimeter and a maximum area. perimeter: 316 meters
1 answer:
Short answer: the rectangle is a square, so the length and width are ...
(316 m)/4 = 79 m.
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Let x represent the length of the rectangle. Then the width in meters is 316/2 -x. The area is the product of length and width.
area = x(158 -x)
This is the equation of a parabola that opens downward. It has zeros at x=0 and x=158. The vertex is on the line of symmetry, halfway between the zeros, at x=79.
The length is 79 m.
The width is 158 -79 = 79 m.
(316 m)/4 = 79 m.
_____
Let x represent the length of the rectangle. Then the width in meters is 316/2 -x. The area is the product of length and width.
area = x(158 -x)
This is the equation of a parabola that opens downward. It has zeros at x=0 and x=158. The vertex is on the line of symmetry, halfway between the zeros, at x=79.
The length is 79 m.
The width is 158 -79 = 79 m.
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