Given a perimeter of 180, how do you find the length and the width of the rectangle of maximum area?

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Given a perimeter of 180, how do you find the length and the width of the rectangle of maximum area?

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Answer 1 · 2016-09-24T23:41:26

Given a perimeter of 180, the length and width of the rectangle with maximum area are 45 and 45.

Explanation:

Let $x =$ $x=$ the length and $y =$ $y=$ the width of the rectangle.
The area of the rectangle $A = x y$ $A =xy$

$2 x + 2 y = 180$ $2x+2y=180$ because the perimeter is $180$ $180$ .

Solve for $y$ $y$
$2 y = 180 - 2 x$ $2y=180-2x$
$y = 90 - x$ $y=90-x$

Substitute for $y$ $y$ in the area equation.
$A = x (90 - x)$ $A=x(90-x)$
$A = 90 x - x^{2}$ $A=90x-x^2$

This equation represents a parabola that opens down. The maximum value of the area is at the vertex.

Rewriting the area equation in the form $a x^{2} + b x + c$ $ax^2+bx+c$
$A = - x^{2} + 90 x a a a a = - 1, b = 90, c = 0$ $A=-x^2+90xcolor(white)(aaa)a=-1, b=90, c=0$

The formula for the $x$ $x$ coordinate of the vertex is
$x = \frac{- b}{2 a} = \frac{- 90}{2 \cdot - 1} = 45$ $x=(-b)/(2a)= (-90)/(2*-1)=45$

The maximum area is found at $x = 45$ $x=45$
and $y = 90 - x = 90 - 45 = 45$ $y=90-x=90-45=45$

Given a perimeter of 180, the dimensions of the rectangle with maximum area are 45x45.

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Given a perimeter of 180, how do you find the length and the width of the rectangle of maximum area?

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