Q. 59 - Calculus

Question

A steam pipe must be buried underground to reach from one corner of a rectangular parking lot to the diagonally opposite corner. The dimensions of the parking lot are 500 feet by 800 feet. It costs 5 dollars per foot to lay steam

pipe under the pavement but only 3 dollars per foot to lay the pipe along one of the long edges of the parking

lot. Because of nearby sidewalks, the pipe cannot be laid along the 500-foot sides of the parking lot. How should

the steam pipe be buried so as to cost as little as possible?

Step-by-Step Solution

Answer

The steam pipe is buried along the 800ft rectangular park is 425ft. 

Step 1.Given Information

Given  rectangular park with 800ft and 500ft

Step 2.Consideration

Assume the length has to be taken has x on the edge of the 800ft

So $CE=x$ then $ED=800-x .$

Using Pythagorean theorem $$\begin{gathered} {\left(ED\right)}^2+{\left(DB\right)}^2={\left(BE\right)}^2 \\ {\left(500\right)}^2+{\left(800-x\right)}^2={\left(BE\right)}^2 \\ 250000+64000+x^2-1600x={\left(BE\right)}^2 \\ 890000+x^2-1600x={\left(BE\right)}^2 \\ \left(BE\right)=\sqrt{890000+x^2-1600x} \end{gathered}$$

Step 3.Cost consideration

The cost buried steam pipe is $$\begin{gathered} C\left(x\right)=3\left(BE\right)+5\left(CE\right) \\ =3x+5\left(\sqrt{890000+x^2-1600x}\right) \end{gathered}$$

Derivative of C(x) on both sides

$$\begin{gathered} \frac{dC\left(x\right)}{dx}=\frac{d}{dx}\left[3\right(x)+5\left(\sqrt{890000+x^2-1600x}\right)] \\ =3+5\left(\frac{1}{2\sqrt{890000+x^2-1600x}}\frac{d }{dx}\right(890000+x^2-1600x\left)\right) \\ =3+5\left(\frac{1}{2\sqrt{890000+x^2-1600x}}\right)(2x-1600) \\ =3+5\left(\frac{1}{\sqrt{890000+x^2-1600x}}\right)(x-800) \end{gathered}$$

Step 4.Substitution

Substitute $$\begin{gathered} \frac{d }{dx}\left(c\right(x\left)\right)=0 \\ 3+5\left(\frac{1}{\sqrt{890000+x^2-1600x}}\right)(x-800)=0 \\ \frac{3\sqrt{890000+x^2-1600x}+5(x-800)}{\sqrt{890000+x^2-1600x}}=0 \\ 3\sqrt{890000+x^2-1600x}+5(x-800)=0 \\ 3\sqrt{890000+x^2-1600x}=-5(x-800) \end{gathered}$$

Squaring on both sides

$$\begin{gathered} {\left[3\sqrt{890000+x^2-1600x}\right]}^2={\left[-5(x-800)\right]}^2 \\ 9(890000+x^2-1600x)=25(x^2+640000-1600x) \\ 801000+9x^2-14400x=25x^2+16000000-40000 \\ 16x^2-25600x+7990000=0 \end{gathered}$$

The values of $x=1175 \& x=425$

the value of rectangular park is 800ft so above is not considered.

So the value of x=425 is the critical point of cost function.

Step 5.To find maxima

Differentiating on both sides $$\begin{gathered} c^{\text{'}}\left(x\right)=3+5\left(\frac{1}{\sqrt{890000+x^2-1600x}}\right(x-800\left)\right) \\ \frac{\partial }{\partial x}\left(c^{\text{'}}\right(x\left)\right)=\frac{d }{dx}(3+5(\frac{1}{\sqrt{890000+x^2-1600x}}(x-800)) \\ =5(\frac{\sqrt{890000+x^2-1600x}{\displaystyle \frac{d }{dx}}(x-800)-(x-800){\displaystyle \frac{d }{dx}}\left(\sqrt{890000+x^2-1600x}\right)}{\sqrt{890000+x^2-1600x}} \\ =5(\frac{\sqrt{890000+x^2-1600x}\left(1\right)-(x-800){\displaystyle \frac{1}{\sqrt{890000+x^2-1600x}}}(x-800)}{\sqrt{890000+x^2-1600x}} \\ =5\left[\frac{890000+x^2-1600x-(x^2+640000-1600x)}{\left(890000+x^2-1600x\right)\sqrt{890000+x^2-1600x}}\right] \\ =5\left[\frac{25000}{\left(890000+x^2-1600x\right)\sqrt{890000+x^2-1600x}}\right] \end{gathered}$$

Step 6.The solution

Substitute x=425 in $c^{\text{'}\text{'}}\left(x\right)$

$$\begin{gathered} c^{\text{'}\text{'}}\left(425\right)=5\left[\frac{250000}{\left(890000+{425}^2-1600\left(425\right)\right)\sqrt{890000+{425}^2-1600\left(425\right)}}\right] \\ {c}^{\text{'}\text{'}}\left(425\right)=5\left[\frac{250000}{\left(390625\right)\sqrt{390625}}\right] \\ {c}^{\text{'}\text{'}}\left(425\right)>0 \end{gathered}$$

The maximum value of x=425

The steam pipe is buried along the 800ft rectangular park is 425ft.