Science:Math Exam Resources/Courses/MATH110/April 2018/Question 10

MATH110 April 2018

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Question 10
You decide to manufacture a can that contains a fixed volume of ${\textstyle 400cm^{3}}$ by bending a rectangular sheet of metal to form a cylinder and then soldering a circular disk at each end so that your can has both a top and a bottom. You use a metal that cost $1 per ${\textstyle cm^{2}}$ for the side and a different metal that costs $2 per ${\textstyle cm^{2}}$ for the top and bottom. Find the radius of the can which will minimize its manufacturing cost. Show that your answer corresponds indeed to an absolute minimum. Recall the volume of a cylinder of height ${\textstyle h}$ and base radius ${\textstyle r}$ is ${\textstyle V=\pi r^{2}h}$ and its surface area is ${\textstyle S=2\pi r^{2}+2\pi rh}$ .

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Hint
Write the total cost as a function of radius and height. Then use the volume constraint to eliminate one of the variables.

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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. We can split the surface area into ${\textstyle S=S_{1}+S_{2}}$ , where ${\textstyle S_{1}}$ is the surface area of the side of the can, and ${\textstyle S_{2}}$ is the area of the top and bottom of the can. Then we have ${\textstyle S_{1}=2\pi rh}$ and ${\textstyle S_{2}=2\pi r^{2}}$ . Now we can calculate the total cost of manufacturing a can by calculating the cost of the sides of the can and the cost of the top and bottom of the can separately. If ${\textstyle C}$ is the total cost in dollars of manufacturing one can, then we have $C=2\pi rh+2\cdot 2\pi r^{2}=2\pi rh+4\pi r^{2},$ since the metal used for the side costs $1 per ${\textstyle cm^{2}}$ and the metal for the top and bottom costs $2 per ${\textstyle cm^{2}}$ . Next, we will make use of the given volume constraint, which can be stated as ${\textstyle 400=\pi r^{2}h}$ . Solving for ${\textstyle h}$ (and noting that we can divide through by ${\textstyle r}$ since ${\textstyle r}$ must be positive), we get $h={\frac {400}{\pi r^{2}}}.$ This will allow us to rewrite the cost function as a function in only one variable. Substituting ${\textstyle h=400/(\pi r^{2})}$ into the cost formula gives $C=2\pi r{\frac {400}{\pi r^{2}}}+4\pi r^{2}={\frac {800}{r}}+4\pi r^{2}.$ To minimize this function, we will find the roots of its first derivative. By the power rule, we get $C'=({\frac {800}{r}}+4\pi r^{2})'=(800r^{-1})'+(4\pi r^{2})'=-800r^{-2}+8\pi r.$ Setting this to zero and rearranging the equation gives $8\pi r={\frac {800}{r^{2}}}.$ Multiplying through by ${\textstyle r^{2}}$ and diving by ${\textstyle 8\pi }$ leaves $r^{3}={\frac {100}{\pi }}.$ It follows that $r={\sqrt[{3}]{\frac {100}{\pi }}}cm.$ Note that the first derivative of the cost function had only one root, so to show that it is an absolute minimum it suffices to show that it is a local minimum (since the endpoints of the domain are not in consideration). Moreover, since the first derivative has no other roots, we can show that the point is a local minimum by simply checking the value of the first derivative of the function at some other point in the domain. We will check the value of the first derivative of ${\textstyle C}$ at the point ${\textstyle r=1}$ (which is smaller than ${\textstyle {\sqrt[{3}]{\frac {100}{\pi }}}}$ ). At ${\textstyle r=1}$ , the first derivative of ${\textstyle C}$ is $8\pi \cdot 1-{\frac {800}{1}}=8\pi -800<0.$ This shows that to the left of the local extremum the function is decreasing, so the radius we found is indeed a local minimum and hence the absolute minimum. Answer: The radius that minimizes the manufacturing cost is ${\textstyle {\color {blue}r={\sqrt[{3}]{\frac {100}{\pi }}}cm}}$ .