Dimensions of a building?

Question

A small office unit is to contain $500 f t^{2}$ $500 ft^2$ of floor space.

(a) express the length $y$ $y$ of the building as a function of the width $x$ $x$

(b) if the walls cost $$ 100$ $$100$ per running foot express the cost $C$ $C$ of the walls as a function of the width $x$ $x$ (Disregard the wall space above the doors and the thickness of the walls.)

I figured out (a) to be $y (x) = \frac{500}{x}$ $y(x) = 500/x$

but im having trouble with (b)

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Answer 1 · 2018-02-18T19:52:12

$C = \frac{300 x^{2} - 600 x + 100000}{x}$ $C=(300x^2-600x+100000)/x$

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Your answer to part $(a)$ $(a)$ is correct.

Part $(b)$ $(b)$ :

The total length of the walls is the perimeter of the building plus the middle divider but we have to subtract the widths of the two doors.

Let the perimeter be $P$ $P$ and total length of the walls be $L$ $L$ .

$P = 2 (x + y)$ $P=2(x+y)$

$L = 2 (x + y) + x - 2 (3) = 2 x + 2 y + x - 6$ $L=2(x+y)+x-2(3)=2x+2y+x-6$

$L = 3 x + 2 y - 6$ $L=3x+2y-6$

Now, we can substitute your answer from part $(a)$ $(a)$ for $y$ $y$ :

$L = 3 x + 2 (\frac{500}{x}) - 6$ $L=3x+2(500/x)-6$

$L = 3 x + \frac{1000}{x} - 6$ $L=3x+1000/x-6$

$C = 100 (3 x + \frac{1000}{x} - 6)$ $C=100(3x+1000/x-6)$

$C = 300 x + \frac{100000}{x} - 600 = \frac{300 x^{2} - 600 x + 100000}{x}$ $C=300x+100000/x-600=(300x^2-600x+100000)/x$