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PrettyInCream's question at Yahoo! Answers regarding minimizing cost to get maximum volume

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MarkFL

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Feb 24, 2012
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Here is the question:

Help with a max/min question regarding a rectangular volume. help please math smarties?


A closed rectangular box with a square base and rectangular sides is to be constructed using two different materials. The bottom and sides are made from stronger material that costs $3, and the top of the box is made from a cheaper materials costing $2 per square foot. If Sasha has $100 to spend on creating this box, what is the box of largest volume she can afford?
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Re: PrettyInCream's question at Yahoo! Amswers regarding minimizing cost to get maximum volume

Hello Sasha,

Let's let all linear measures be in feet and the cost be in dollars. First we need to find our objective function, which is the cost function $C$. Let's let $x$ be the length of the sides of the square base, and $h$ be the height of the box, where $h,x>0$. And since cost is cost per area times area, we may state:

\(\displaystyle C(h,x)=3\left(x^2+4hx \right)+2\left(x^2 \right)\)

Now, we are constrained by the relationship between $x$, $h$ and the volume of the box $v$:

\(\displaystyle V=hx^2\)

From the constraint, we find when solving for $h$:

\(\displaystyle h=\frac{V}{x^2}\)

And so we may now express the cost function in terms of the single variable $x$ by substituting for $h$ into the objective function:

\(\displaystyle C(x)=3\left(x^2+\frac{4V}{x} \right)+2\left(x^2 \right)=5x^2+12Vx^{-1}\)

Equating the first derivative to zero, we may find the critical value:

\(\displaystyle C'(x)=10x-12Vx^{-2}=\frac{2\left(5x^3-6V \right)}{x^2}=0\)

This implies:

\(\displaystyle 5x^3-6V=0\)

\(\displaystyle x=\left(\frac{6V}{5} \right)^{\frac{1}{3}}\)

Now, to determine the nature of the extremum associated with this critical value, we may use the second derivative test.

\(\displaystyle C''(x)=10+24Vx^{-3}\)

We see that for all $x>0$ we have $C''(x)>0$ which means our extremum is a minimum. Hence:

\(\displaystyle C_{\min}=C\left(\left(\frac{6V}{5} \right)^{\frac{1}{3}} \right)=5\left(\left(\frac{6V}{5} \right)^{\frac{1}{3}} \right)^2+12V\left(\left(\frac{6V}{5} \right)^{\frac{1}{3}} \right)^{-1}=3\left(180V^2 \right)^{\frac{1}{3}}\)

Since Sasha has \$100 to spend on the box, equating this minimized cost function to the amount she can spend will have the effect of maximizing the volume she can get for her money:

\(\displaystyle 100=3\left(180V^2 \right)^{\frac{1}{3}}\)

Now we want to solve for $V$. Divide through by 3:

\(\displaystyle \frac{100}{3}=\left(180V^2 \right)^{\frac{1}{3}}\)

Cube both sides:

\(\displaystyle \left(\frac{100}{3} \right)^3=180V^2\)

Divide through by 180:

\(\displaystyle V^2=\frac{1}{180}\left(\frac{100}{3} \right)^3\)

Take the positive root to find the volume in cubic feet:

\(\displaystyle V=\frac{1}{6\sqrt{5}}\left(\frac{100}{3} \right)^{\frac{3}{2}}=\frac{100}{9}\sqrt{\frac{5}{3}}\approx14.3443827637312\)