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

• MHB
• MarkFL
In summary: Smile)In summary, the problem involves finding the height and radius of a cone-shaped paper drinking cup that holds 10ml of water with the least amount of paper. Using Lagrange multipliers, we can derive equations and solve for the optimal values of height and radius. Alternatively, we can also use differentiation and substitution to find the optimal values. The result is that the optimal height and radius are both equal to the cubic root of 6V/pi where V is the volume of water in ml.
MarkFL
Gold Member
MHB
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.

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 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.

## What is optimization with constraint?

Optimization with constraint is a mathematical process that involves finding the best possible solution for a problem while adhering to certain limitations or restrictions.

## What are some common examples of optimization with constraint?

Some common examples of optimization with constraint include maximizing profits while minimizing costs, finding the shortest route for a delivery while considering traffic and time constraints, and maximizing efficiency in resource allocation.

## How is optimization with constraint different from regular optimization?

Regular optimization involves finding the best solution for a problem without any limitations or restrictions. Optimization with constraint, on the other hand, takes into account the constraints and finds the best possible solution within those constraints.

## What are the steps involved in optimization with constraint?

The steps involved in optimization with constraint typically include defining the problem, identifying the constraints, formulating a mathematical model, solving the model using appropriate techniques, and interpreting the results to make decisions.

## What are some common techniques used for optimization with constraint?

Some common techniques used for optimization with constraint include linear programming, dynamic programming, and heuristic algorithms. The choice of technique depends on the problem at hand and the available resources.

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