Single Variable Calculus

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Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 12

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Find the dimensions of the box that minimize the amount of material used. Suppose that the box is made from square base and has open top with volume of 32,000cm$^3$

$ \begin{equation} \begin{aligned} \text{Volume = } x(x)(y) &= x^2y = 32000\text{cm}^3\\ \\ x^y &= 32000\\ \\ y &= \frac{32000}{x^2} \quad \Longleftarrow \text{(Equation 1)} \end{aligned} \end{equation} $

If the equation asked is to minimize the amount of material used, it means that we have to minimize the total surface area of the box...

Recall that the total surface area $A$ is the same of the area of each of the faces of the box. So, $A = x^2 + xy + xy + xy + xy = x^2 + 4xy$

$A = x^2 + 4xy$ (remember that the box has open top, its area is not included)

If we substitute Equation 1,

$\displaystyle A = x^2 + 4x \left( \frac{32000}{x^2} \right) = x^2 + \frac{128000}{x}$

If we set $A' = 0$

$ \begin{equation} \begin{aligned} A' = 0 &= 2x - \frac{128000}{x^2}\\ \\ \frac{128000}{x^2} & = 2x\\ \\ x^3 & = \frac{128000}{2}\\ \\ x &= \sqrt[3]{\frac{128000}{2}}\\ \\ x &= 40 \text{ cm}\\ \\ \text{if } x &= 40, \text{ then}\\ \\ y &= \frac{32000}{40^2}\\ \\ y &= 20\text{ cm} \end{aligned} \end{equation} $

Therefore, the dimension of the box. That would minimize the amount of material would be $40 \times 40 \times 20$cm

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