Answer to Question #148341 in Calculus for Angelo

Question

A one-story building having a rectangular floor space of 13,200 𝑓𝑡^2 is to be constructed

where a 22 ft walkway is required in the front and back, and a 15 ft walkway is required on each side. Find the dimensions of the lot having the least area on which this building can be located.

Answer 1

Consider the rectangular floor of length $x$ and width $y$.

Given, $xy=13200$ , so $y=\frac{13200}{x}$

The area of the lot on which the building is to be constructed is,

$A=(x+30)(y+44)$

Substitute $y=\frac{13200}{x}$ to express the area in a single variable $x$ as,

$A(x)=(x+30)(\frac{13200}{x}+44)$

$=13200+44x+\frac{396000}{x}+1320$

$=14520+44x+\frac{396000}{x}$

Differentiate $A(x)$ with respect to $x$ and set it equal to zero in order to find the critical value as,

$A'(x)=0$

$44-\frac{396000}{x^2}=0$

$x^2=9000$

$x=30\sqrt{10}$

Here, "A"(x)=\\frac{792000}{x^3}>0" for $x=30\sqrt{10}$ , so $A$ is minimum at $x=30\sqrt{10}$

Therefore, the dimensions of the floor are:$x=30\sqrt{10}\approx94.87$ ft$y=\frac{13200}{30\sqrt{10}}=44\sqrt{10}\approx139.14$ ftThe dimensions of the lot are:
$x+30=30\sqrt{10}+30\approx124.87 ft$
$y+44=44\sqrt{10}+44\approx 183.14 ft$