Eggy's question at Yahoo Answers regarding optimization with constraint

[MarkFL]

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.

 

[MarkFL]

Hello Eggy,

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

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

subject to the constraint on the volume in ml:

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

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:

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

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

This system may be simplified to:

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

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

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

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

This implies:

$$h^2=2r^2$$

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

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

Solving for $h$ we obtain:

$$h=\sqrt[3]{\frac{6V}{\pi}}$$

and so:

$$r=\frac{h}{\sqrt{2}}=\frac{\sqrt[3]{\frac{6V}{\pi}}}{\sqrt{2}}$$

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

$$h=\sqrt[3]{\frac{60}{\pi}}$$

$$r=\frac{\sqrt[3]{\frac{60}{\pi}}}{\sqrt{2}}$$

To Eggy and any other guests viewing this topic, I invite and encourage you to post other optimization problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.

 

[Unknown008]

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

A=\pi r\sqrt{r^2+h^2}

V=\dfrac{\pi}{3}hr^2

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.

V=\dfrac{\pi}{3}hr^2

h = \dfrac{3V}{\pi r^2}

Substituting in the area expression:

A=\pi r\sqrt{r^2+\dfrac{9V^2}{\pi^2 r^4}}

Or we could write it like this:

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

A=\sqrt{\pi^2r^4 + \dfrac{9V^2}{r^2}}

Differentiate and set to zero for optimum:

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

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

2\pi^2r^3 - \dfrac{9V^2}{r^3} = 0

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

The height is then:

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

Simplified to:

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

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.