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

AI Thread Summary
The discussion centers on optimizing the dimensions of a cone-shaped paper cup to minimize material while holding a fixed volume of 10 mL. Using Lagrange multipliers, the optimal height and radius are derived, leading to the equations h^2 = 2r^2 and specific formulas for h and r in terms of the volume. An alternative method is presented that involves differentiating the surface area and substituting the volume constraint, ultimately arriving at the same results for height and radius. Both approaches confirm that the height is h = (6V/π)^(1/3) and the radius is r = h/√2. The thread encourages further optimization problem discussions in the forum.
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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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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.
 
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.
 
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