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torres' question at Yahoo! Answers regarding minimizing the cost of a box
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Hello torres,
If we let $x$ be the side length of the bottom of the box and $h$ be the height of the box, then the constant volume V of the box is:
\(\displaystyle V=hx^2\)
This is our constraint.
If we then let $c_1$ be the cost per unit area of the top/bottom and $c_2$ be the cost per unit area of the sides, then our total cost function $C$ is:
\(\displaystyle C(h,x)=2c_1x^2+4c_2hx\)
This is our objective function.
Now, if we solve the constraint for $h$, we obtain:
\(\displaystyle h=\frac{V}{x^2}\)
Substituting for $h$ into the objective function, we obtain:
\(\displaystyle C(x)=2c_1x^2+4c_2Vx^{-1}\)
Now, to find the critical value(s), we should differentiate with respect to $x$ and equate the result to zero:
\(\displaystyle C'(x)=4c_1x-4c_2Vx^{-2}=\frac{4\left(c_1x^3-c_2V \right)}{x^2}=0\)
This implies:
\(\displaystyle c_1x^3-c_2V=0\implies x=\left(\frac{c_2}{c_1}V \right)^{\frac{1}{3}}\)
We can see that for values of $x$ less than this critical value the first derivative is negative and for values larger than this critical value the first derivative is positive, therefore by the first derivative test we may conclude that this critical value is at a relative minimum.
Thus the height of the box at this critical value is:
\(\displaystyle h=\frac{V}{\left(\frac{c_2}{c_1}V \right)^{\frac{2}{3}}}=\left(\left(\frac{c_1}{c_2} \right)^2V \right)^{\frac{1}{3}}=\frac{c_1}{c_2}x\)
Now, plugging in the given data:
\(\displaystyle V=6000\text{ in}^3,\,c_1=5\frac{\text{¢}}{\text{in}^2},\, c_2=3\frac{\text{¢}}{\text{in}^2}\)
We find:
\(\displaystyle x=\left(\frac{3}{5}\cdot6000 \right)^{\frac{1}{3}}\text{ in}=2\sqrt[3]{450}\text{ in}\)
\(\displaystyle h=\frac{5}{3}\cdot2\sqrt[3]{30}\text{ in}=\frac{10}{3}\sqrt[3]{450}\text{ in}\)
If we let $x$ be the side length of the bottom of the box and $h$ be the height of the box, then the constant volume V of the box is:
\(\displaystyle V=hx^2\)
This is our constraint.
If we then let $c_1$ be the cost per unit area of the top/bottom and $c_2$ be the cost per unit area of the sides, then our total cost function $C$ is:
\(\displaystyle C(h,x)=2c_1x^2+4c_2hx\)
This is our objective function.
Now, if we solve the constraint for $h$, we obtain:
\(\displaystyle h=\frac{V}{x^2}\)
Substituting for $h$ into the objective function, we obtain:
\(\displaystyle C(x)=2c_1x^2+4c_2Vx^{-1}\)
Now, to find the critical value(s), we should differentiate with respect to $x$ and equate the result to zero:
\(\displaystyle C'(x)=4c_1x-4c_2Vx^{-2}=\frac{4\left(c_1x^3-c_2V \right)}{x^2}=0\)
This implies:
\(\displaystyle c_1x^3-c_2V=0\implies x=\left(\frac{c_2}{c_1}V \right)^{\frac{1}{3}}\)
We can see that for values of $x$ less than this critical value the first derivative is negative and for values larger than this critical value the first derivative is positive, therefore by the first derivative test we may conclude that this critical value is at a relative minimum.
Thus the height of the box at this critical value is:
\(\displaystyle h=\frac{V}{\left(\frac{c_2}{c_1}V \right)^{\frac{2}{3}}}=\left(\left(\frac{c_1}{c_2} \right)^2V \right)^{\frac{1}{3}}=\frac{c_1}{c_2}x\)
Now, plugging in the given data:
\(\displaystyle V=6000\text{ in}^3,\,c_1=5\frac{\text{¢}}{\text{in}^2},\, c_2=3\frac{\text{¢}}{\text{in}^2}\)
We find:
\(\displaystyle x=\left(\frac{3}{5}\cdot6000 \right)^{\frac{1}{3}}\text{ in}=2\sqrt[3]{450}\text{ in}\)
\(\displaystyle h=\frac{5}{3}\cdot2\sqrt[3]{30}\text{ in}=\frac{10}{3}\sqrt[3]{450}\text{ in}\)