Answer to Question #198831 in Algebra for Zainab

Question

A small rectangular warehouse is to be constructed which is to have an area of 10000 square feet. The building is to be partitioning internally in to eight equal parts. The costs have been estimated based on exterior and interior walls dimensions. The costs are $300 per running foot of exterior wall and plus $1100 per running foot of interior wall.

a) Determine the dimensions which will minimizes the construction costs?

b) What are the minimum cost?

Answer 1

a)Area=x.y

$10000=x.y$

$Y=\frac {10000} {x}$

Perimeter=3x+5y

Perimeter$=3×\frac {10^4} {x} +5x$

Cost$=300(2x+2y)+1100(3x+y)$

(300 for exterior walls and 1100 for interior walls.)

Cost $=300×2(x+\frac {10000} {x} )+1100(3x+\frac {10000} {x} )$

To minimize Cost= d/dx(cost)=0. Find x

$\frac {D} {dx} {600(x+10000^{-1} )+1100(3x+10000^{-1} )}=0$

$600(1+1000(-1)x^{-2} )+1100(3+10000(-1)x^{-2})=0$

$6-\frac {6000} {x^2} +33-\frac {110000} {x^2} =0$

$39x^2-170000=0$

$X^2=\frac {170000} {39}$

X=66.02 and Y=151.4

b)Put x=66.02 in an equation to find minimum cost.

Cost=600(66.02+10000/66.02)+1100(198.06+10000/66.02)

Cost=(600×217.98) +(1100×349.52)

Cost=130,488+384,472

Cost=$514,960