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Eggy's question at Yahoo! Answers regarding optimization with constraint

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Feb 24, 2012
Here is the question:

Calculus Max Min problem help?

A cone-shaped paper drinking cup is to hold 10ml of water. What is height and radius of the cup that will require the least amount of paper?
Here is a link to the question:

Calculus Max Min problem help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
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Feb 24, 2012
Hello Eggy,

Using Lagrange multipliers, we have the objective function (using the formula for the lateral surface area of a cone):

\(\displaystyle f(h,r)=\pi r\sqrt{r^2+h^2}\)

subject to the constraint on the volume in ml:

\(\displaystyle g(h,r)=\frac{\pi}{3}hr^2-V=0\)

We have used the constant $V$ rather than the given value as we can just plug this in at the end of the problem.

Hence, we obtain the system:

\(\displaystyle \pi r\frac{h}{\sqrt{r^2+h^2}}=\lambda\left(\frac{\pi r^2}{3} \right)\)

\(\displaystyle \pi\left(r\frac{r}{\sqrt{r^2+h^2}}+\sqrt{r^2+h^2} \right)=\lambda\left(\frac{2\pi hr}{3} \right)\)

This system may be simplified to:

\(\displaystyle \frac{h}{\sqrt{r^2+h^2}}=\lambda\left(\frac{r}{3} \right)\)

\(\displaystyle \frac{2r^2+h^2}{\sqrt{r^2+h^2}}=\lambda\left(\frac{2hr}{3} \right)\)

Solving both equation for $\lambda$ and equating, we obtain:

\(\displaystyle \lambda=\frac{3h}{r\sqrt{r^2+h^2}}=\frac{3(2r^2+h^2)}{2hr\sqrt{r^2+r^2}}\)

This implies:

\(\displaystyle h^2=2r^2\)

Substituting into the constraint for $r^2$, we find:

\(\displaystyle \frac{\pi}{3}h\left(\frac{h^2}{2} \right)-V=0\)

Solving for $h$ we obtain:

\(\displaystyle h=\sqrt[3]{\frac{6V}{\pi}}\)

and so:

\(\displaystyle r=\frac{h}{\sqrt{2}}=\frac{\sqrt[3]{\frac{6V}{\pi}}}{\sqrt{2}}\)

Now, since \(\displaystyle V=10\text{ mL}\), and \(\displaystyle 1\text{ mL}=1\text{ cm}^3\), we find that $r$ and $h$ in cm are:

\(\displaystyle h=\sqrt[3]{\frac{60}{\pi}}\)

\(\displaystyle r=\frac{\sqrt[3]{\frac{60}{\pi}}}{\sqrt{2}}\)

To Eggy and any other guests viewing this topic, I invite and encourage you to post other optimization problems in our Calculus forum.

Best Regards,



Feb 1, 2012
Well, since I don't know about Lagrange Multipliers, this is how I would have done it:

[tex]A=\pi r\sqrt{r^2+h^2}[/tex]


Keeping in mind that V is a constant (in this case 10).

Since we're minimizing the area, we need to differentiate A, but having r and h as variables, we need to use the expression for V to get h in terms or r, or r in terms of h. I'm going with h in terms of r.


[tex]h = \dfrac{3V}{\pi r^2}[/tex]

Substituting in the area expression:

[tex]A=\pi r\sqrt{r^2+\dfrac{9V^2}{\pi^2 r^4}}[/tex]

Or we could write it like this:

[tex]A=\sqrt{\pi^2r^2\left(r^2+\dfrac{9V^2}{\pi^2 r^4}\right)}[/tex]

[tex]A=\sqrt{\pi^2r^4 + \dfrac{9V^2}{r^2}}[/tex]

Differentiate and set to zero for optimum:

[tex]A'=\dfrac12 \left(\pi^2r^4 + \dfrac{9V^2}{r^2}\right)^{-0.5} \cdot \left(4\pi^2r^3 - \dfrac{18V^2}{r^3}\right) = 0[/tex]

[tex]\dfrac{\left(2\pi^2r^3 - \dfrac{9V^2}{r^3}\right)}{\sqrt{\pi^2r^4 + \dfrac{9V^2}{r^2}}}=0[/tex]

[tex]2\pi^2r^3 - \dfrac{9V^2}{r^3} = 0[/tex]

[tex]r = \sqrt[3]{\dfrac{3V}{\sqrt{2}\pi}}[/tex]

The height is then:

[tex]h = \dfrac{3V}{\pi \left(\sqrt[3]{\dfrac{3V}{\sqrt{2}\pi}}\right)^2}[/tex]

Simplified to:

[tex]h = \sqrt[3]{\dfrac{6V}{\pi}}[/tex]

Which are both the same as what Mark obtained. (Happy)

My method just maybe is a little longer/tedious because of many simplifications involved, but that's using the tools I know.