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Question #e2dc4

1 Answer

Feb 7, 2018

Length: $x \geq 16$ $x>=16$ , Width: $y \geq 32$ $y>=32$

Explanation:

I'm not understanding this too well, because there is no $x$ $x$ variable mentioned above. Anyway, let's give it a try.

Since an area of a rectangle can't be negative, it has to be the absolute value of something. Let's try this.

$4 x - 64 \geq 0$ $4x-64>=0$
$x \geq 16$ $x>=16$

Since we found the length of the rectangle, the width would be $x + 16 \geq 32$ $x+16>=32$ .

Yay!

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