This is just a question for my research on the ideal cup size to minimize the amount of plastic used.
Assuming that the conical frustum is in a shape of a cup. How do I go about in solving the maximum volume of this cup using calculus to minimize the amount of plastic? I stumbled upon some sort of confusion between integration and optimization, can someone explain to me what would be the best method in approaching this? (And please involve calculus, thanks!)
Let the upside down cup be a frustum formed from a cone with base radius $b$ and height $h=cb$, where $c$ is a dimensionless constant (the cone's slope). The upper radius is $a<b$.
Then the volume is given by subtracting the volume of two cones, each a third of base area times height, $$V=\frac{\pi}{3}c(b^3-a^3).$$ The surface area is also obtained by subtracting the area of two cones, each equal to $\pi rl$, and adding the top disc (assuming the cup has a wider mouth than base): $$A=\pi (b^2-a^2)\sqrt{1+c^2}+\pi a^2$$
The question asks to minimize $A$ given $V$. Without loss of generality, we can ignore the factor of $\pi$ in the formula for $A$, and take $c(b^3-a^3)=v=3V/\pi$ as the constraint.
Using the method of Lagrange multipliers, let $$f(a,b,c):=\sqrt{1+c^2}(b^2-a^2)+a^2-\lambda c(b^3-a^3)$$ Then differentiating with respect to $a,b,c$, we get $\nabla f=0$, \begin{align} 2\sqrt{1+c^2}a-2a&=3\lambda ca^2\\ 2\sqrt{1+c^2}b&=3\lambda cb^2\\ c(b^2-a^2)&=\lambda\sqrt{1+c^2}(b^3-a^3)\\ c(b^3-a^3)&=v \end{align}
(i) Solving these (with computing help), gives, assuming $a\ne0$, $$a=\sqrt[6]{\frac{44}{63 \sqrt{7}}-\frac{16}{63}} x\approx0.464x,\quad b=\sqrt[6]{\frac{13}{126}+\frac{44}{63 \sqrt{7}}} x\approx0.846x,\quad c=\sqrt{\frac{14+8 \sqrt{7}}{9}}\approx1.977,\quad \lambda= \frac{2^{2/3} \sqrt[3]{2+\sqrt{7}}}{3 x}$$ where $x=v^{1/3}$. The height of the cup is $c(b-a)\approx0.755x$.
The volume is then $\frac{\pi v}{3}=V$ while the surface area is $$A=\sqrt[3]{\frac{2+\sqrt{7}}{2}} \pi x^2\approx4.16x^2\approx4.03V^{2/3}$$
(ii) The alternative is $a=0$, for which the cup becomes an open cone. Then $$V=\frac{\pi}{3}cb^3,\qquad A=\pi b^2\sqrt{1+c^2}$$ A straightforward minimization of $A$ keeping $V$ fixed, gives $$0=\frac{dA}{db}=\frac{d}{db}\pi b^2\sqrt{1+v^2/b^6}\iff b=\left(\frac{3}{\pi\sqrt2}\right)^{1/3}V^{1/3}$$ $$A=\pi b^2\sqrt{1+c^2}=3^{7/6}\left(\frac{\pi}{2}\right)^{1/3}V^{2/3}\approx4.188V^{2/3}$$
So it is option (i) that minimizes the area.
The shape looks like this: