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) Let $x=$ the length of warehouse in $ft,$ $y=$ the width ofwarehouse in $ft.$

Then $xy=1000\ ft^2$

Solve for $x$

$$x=\dfrac{10000}{y}, y>0$$

The construction cost is

$$C=2x(300)+2y(300)+7y(1100)$$

Substitute

$$C=C(y)=\dfrac{6000000}{y}+8300y, y>0$$

Find the first derivative with respect to $y$

$$C'(y)=(\dfrac{6000000}{y}+8300y)'$$

$$=-\dfrac{6000000}{y^2}+8300$$

Find the critical number(s)

$$C'(y)=0=>-\dfrac{6000000}{y^2}+8300=0$$

$$y^2=\dfrac{60000}{83}$$

$$y_1=-100\sqrt{\dfrac{6}{83}}\approx-26.8866,$$

$$y_2=100\sqrt{\dfrac{6}{83}}\approx26.8866$$

The critical numbers $-100\sqrt{\dfrac{6}{83}}\approx26.8866,100\sqrt{\dfrac{6}{83}}\approx26.8866$

Since $y>0,$ then we consider $100\sqrt{\dfrac{6}{83}}\approx26.8866$

If $0<y<100\sqrt{\dfrac{6}{83}}\approx26.8866, C'(y)<0, C(y)$ decreses.

If $y>100\sqrt{\dfrac{6}{83}}\approx26.8866, C'(y)>0, C(y)$ increses.

The function $C$ has a local minmum at $y=100\sqrt{\dfrac{6}{83}}\approx26.8866.$

Since the function $C$ has the only extremum for $y>0,$ then the function $C$ has the absolute minimum at $y=100\sqrt{\dfrac{6}{83}}\approx26.8866$ for $y> 0.$

$$x=\dfrac{10000}{y}=100\sqrt{\dfrac{83}{6}}\approx371.9319$$

The cost will be minimum if

$$length=100\sqrt{\dfrac{83}{6}}\ ft\approx371.9319\ ft$$

$$width=100\sqrt{\dfrac{6}{83}}\ ft\approx26.8866\ ft$$

b)

$$C(100\sqrt{\dfrac{6}{83}})=600(100\sqrt{\dfrac{83}{6}})+8300(100\sqrt{\dfrac{6}{83}})$$

$$=\$446,318.27$$

The minimum cost is $\$446,318.27.$