Science:Math Exam Resources/Courses/MATH104/December 2011/Question 03

MATH104 December 2011

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[hide] Question 03
A box-shaped shipping crate with a square base is designed to have a volume of 16 m $^{3}$ . The material used to make the base costs twice as much per square metre as the material on the sides. The material on the top costs half as much as the material in the sides. Suppose the material to make the sides costs $20 per square metre. What are the dimensions of the crate that minimize the cost of the materials?

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[show] Hint
What is the function we're trying to optimize? Are there any constraints to consider?

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[show] Solution
Let's call the dimension of the length and width of the base $x$ (they are the same since it is a square base. Let's call the height of the box $y$ . We want to find the dimensions of a box (x and y) that minimizes the cost of building it but such that it has a prescribed volume (16m $^{3}$ ). The volume requirement is known as our constraint and it allows us to find a relationship between our two variables so that we have only one independent variable. The volume is $\displaystyle {}V=x^{2}y=16$ and so we conclude that $y={\frac {16}{x^{2}}}.$ We now need to consider the cost. The total cost will be the sum of building the 4 sides ( $C_{S}$ ), the base ( $C_{B}$ ), and the top ( $C_{T}$ ), $\displaystyle {}C=4C_{S}+C_{B}+C_{T}.$ We have that the cost per square metre of the sides is $20 but for now let's call this number $m$ . We then have that $\displaystyle {}C_{S}=mA_{S}$ where $A_{S}$ is the area of the sides. We know the top costs half as much per square metre and so, $\displaystyle {}C_{T}={\frac {m}{2}}A_{T}.$ Similarly the base costs twice as much per square metre, $\displaystyle {}C_{B}=2mA_{B}.$ where $A_{B}$ and $A_{T}$ are the base and top areas respectively. In order to continue, we need to write each surface area in terms of our independent variable $x$ . The surface area of both the base and the top is $\displaystyle {}A_{B}=A_{T}=x^{2}.$ The surface area of each of the sides is $A_{S}=xy=x{\frac {16}{x^{2}}}={\frac {16}{x}}$ where we have used our relationship for $y$ in terms of $x$ . We can now write the cost ${\begin{aligned}C&=4C_{S}+C_{B}+C_{T}\\&=4m{\frac {16}{x}}+2mx^{2}+{\frac {m}{2}}x^{2}\\&=m\left({\frac {64}{x}}+{\frac {5}{2}}x^{2}\right).\end{aligned}}$ We seek to find critical points of this function and therefore we take its derivative and set it to zero, ${\frac {{\textrm {d}}C}{{\textrm {d}}x}}=m\left(-{\frac {64}{x^{2}}}+5x\right)=0$ and we see that the only critical point is $x^{3}={\frac {64}{5}}.$ Before continuing, it is worth noting that the cost per square metre of the sides (m=$20) was irrelevant information. In order to test if our critical point is indeed a minimum we will use the second derivative test (since there is no closed interval we cannot test the endpoints). The second derivative is, ${\frac {{\textrm {d}}^{2}C}{{\textrm {d}}x^{2}}}=m\left({\frac {128}{x^{3}}}+5\right).$ If we sub in our critical point, ${\frac {{\textrm {d}}^{2}C}{{\textrm {d}}x^{2}}}=m\left(128{\frac {5}{64}}+5\right)=m(10+5)=15m=300$ where here we have used the fact that m=20. Since the second derivative is positive, the objective function, $C$ is concave up and thus we have a minimum at our critical point. Therefore the dimensions that minimize the cost of producing the box with volume 16m $^{3}$ is $x=\left({\frac {64}{5}}\right)^{1/3}\quad y={\frac {16}{x^{2}}}=5^{2/3}.$