Science:Math Exam Resources/Courses/MATH100/December 2015/Question 09

MATH100 December 2015

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[hide] Question 09
A business wants to manufacture a huge closed box with a square base, a square top, and rectangular sides. The material used for the four upright sides of the box costs $2 per square metre, and the material used for the base and top of the box costs $8 per square metre. Find the dimensions of the box with lowest possible cost that has volume 32 m3.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Let h and x be the height of the box and the side length of the square base, respectively. How do you write the volume of the box and its cost as functions of these variable, x and h?

[show] Hint 2
Write the total cost function in terms of $x$ and $h$. You would like to minimize this function taking into account that you have one auxiliary formula given by the volume.

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[show] Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. As we mentioned in the Hint, let h and x be the height of the box and the side length of the square base, respectively. By its definition, we only consider positive x and h. Then, the volume V of the box is $V(x,h)=x^{2}h$ and the cost C for the box can be expressed as ${\begin{aligned}C(x,h)&={\text{cost of the top and the base}}+{\text{cost of the sides}}\\&=8\cdot {\text{area of the top and the base}}+2\cdot {\text{area of the sides}}\\&=8\cdot 2x^{2}+2\cdot 4xh=16x^{2}+8xh.\end{aligned}}$ Since the volume is fixed as $32m^{3}$, we have $x^{2}h=32$. Plugging $h={\frac {32}{x^{2}}}$ into the cost function C, we have $C(x,{\frac {32}{x^{2}}})=16x^{2}+8x\cdot {\frac {32}{x^{2}}}=16x^{2}+{\frac {256}{x}}=f(x)$. Therefore to minimize the cost, it is enough to minimize the function $f$ on the domain $\{x>0\}$. From its derivative $f'(x)=32x-{\frac {256}{x^{2}}}={\frac {32}{x^{2}}}(x^{3}-8)={\frac {32}{x^{2}}}(x-2)(x^{2}+2x+4)$, the function $f$ has one critical point $x=2$. Indeed, we use $x^{2}+2x+4=(x+1)^{2}+3>0$ for any $x>0$. Furthermore, $f'$ changes its sign from - to + at the critical point, we obtain the minimum value of the cost $f(x)$ at $x=2$. Recall that the relation between x and h, $h={\frac {32}{x^{2}}}$. Then, we minimize the cost of the box when the box has the side length and height of $\color {blue}{x=2m,h=8m}$ respectively.