# Optimization questions

#### momopeaches

In my calculus 1 class, we're covering optimization. I was doing fine on the problems until I got sick for a few days, and missed the beginning of finding the minimum and maximum of areas and perimeters for different shapes. So far, I'm stuck on a few problems.

1.
If 1400 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

I'm not quite sure how to maximize volume given perimeter, but I know how to find perimeter given a volume. I'm sure its something simple that I'm over looking.

2.
A cone-shaped cup is made from a circular piece of paper of radius 20 by cutting out a sector and joining the edges AC and BC.

Find the maximum capacity (volume) of such a cup.

Essentially, given a radius of 20, what is the maximum volume that the cone can have. All of the other problems we've covered have involved squares or boxes. I don't even know how to start on this one.

I would really appreciate any help. Between being sick and working, I've having a hard time keeping up, especially when our book has no examples of how to do these kinds of problems.

#### skeeter

MHF Helper
In my calculus 1 class, we're covering optimization. I was doing fine on the problems until I got sick for a few days, and missed the beginning of finding the minimum and maximum of areas and perimeters for different shapes. So far, I'm stuck on a few problems.

1.
If 1400 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

I'm not quite sure how to maximize volume given perimeter, but I know how to find perimeter given a volume. I'm sure its something simple that I'm over looking.

2.
A cone-shaped cup is made from a circular piece of paper of radius 20 by cutting out a sector and joining the edges AC and BC.

Find the maximum capacity (volume) of such a cup.

Essentially, given a radius of 20, what is the maximum volume that the cone can have. All of the other problems we've covered have involved squares or boxes. I don't even know how to start on this one.

I would really appreciate any help. Between being sick and working, I've having a hard time keeping up, especially when our book has no examples of how to do these kinds of problems.
for your first problem, go to the link and scroll down to problem #3

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxmindirectory/MaxMin.html

#### skeeter

MHF Helper
the second problem should not be given to beginners. hope you can follow this ...

let $$\displaystyle R$$ = large circle radius (that's the 20 ... I'll use R for generality)

$$\displaystyle r$$ = radius of the cone's base

$$\displaystyle h$$ = cone's height

$$\displaystyle \theta$$ = angle of the cut out sector (in radians)

note that radius of the circle, $$\displaystyle R$$ = slant height of the cone.

$$\displaystyle h^2 = R^2 - r^2$$

$$\displaystyle h = \sqrt{R^2 - r^2}$$

$$\displaystyle \textcolor{red}{\frac{dh}{dr} = -\frac{r}{\sqrt{R^2-r^2}}}$$

$$\displaystyle V = \frac{\pi}{3} r^2 h$$

$$\displaystyle r^2 = R^2 - h^2$$ substitute for $$\displaystyle r^2$$ ...

$$\displaystyle V = \frac{\pi}{3}(R^2-h^2)h = \frac{\pi}{3}(R^2h-h^3)$$

$$\displaystyle \textcolor{red}{\frac{dV}{dh} = \frac{\pi}{3}(R^2 - 3h^2)}$$

circumference of cone's base $$\displaystyle = 2\pi r = 2\pi R - R\theta$$

$$\displaystyle r = R - \frac{R}{2\pi} \theta$$

$$\displaystyle \textcolor{red}{\frac{dr}{d\theta} = -\frac{R}{2\pi}}$$

from the chain rule ...

$$\displaystyle \frac{dV}{d\theta} = \textcolor{red}{\frac{dV}{dh} \cdot \frac{dh}{dr} \cdot \frac{dr}{d\theta}}$$

$$\displaystyle \frac{dV}{d\theta} = \frac{\pi}{3}(R^2-3h^2) \cdot \left(-\frac{r}{\sqrt{R^2-r^2}}\right) \cdot \left(-\frac{R}{2\pi}\right)$$

setting $$\displaystyle \frac{dV}{d\theta} = 0$$ , note that the only factor than can equal 0 is $$\displaystyle (R^2 - 3h^2)$$ ...

$$\displaystyle h^2 = \frac{R^2}{3}$$

$$\displaystyle r = \sqrt{R^2 - h^2} = \sqrt{R^2 - \frac{R^2}{3}} = \sqrt{\frac{2R^2}{3}}$$

finally ...

$$\displaystyle r = R - \frac{R}{2\pi} \theta$$

$$\displaystyle \sqrt{\frac{2R^2}{3}} = R - \frac{R}{2\pi} \theta$$

solving for $$\displaystyle \theta$$ ...

$$\displaystyle \sqrt{\frac{2}{3}} \cdot R = R\left(1 - \frac{\theta}{2\pi}\right)$$

$$\displaystyle \sqrt{\frac{2}{3}} = 1 - \frac{\theta}{2\pi}$$

$$\displaystyle \frac{\theta}{2\pi} = 1 - \sqrt{\frac{2}{3}}$$

$$\displaystyle \theta = 2\pi\left(1 - \sqrt{\frac{2}{3}}\right) \approx 1.153 \, rad \, \approx 66^\circ$$

• momopeaches

#### HallsofIvy

MHF Helper
In my calculus 1 class, we're covering optimization. I was doing fine on the problems until I got sick for a few days, and missed the beginning of finding the minimum and maximum of areas and perimeters for different shapes. So far, I'm stuck on a few problems.

1.
If 1400 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

I'm not quite sure how to maximize volume given perimeter, but I know how to find perimeter given a volume. I'm sure its something simple that I'm over looking.
Perhaps you are overlooking the fact that there is NO "perimeter". You are given the condition that surface area is fixed. In the length of a side of the square base is "x" and the height is "y", the volume is $$\displaystyle x^2y$$ and the surface area is $$\displaystyle x^2+ 4xy= 1400$$. Solve that for y and replace y in the volume with that formula.

2.
A cone-shaped cup is made from a circular piece of paper of radius 20 by cutting out a sector and joining the edges AC and BC.

Find the maximum capacity (volume) of such a cup.

Essentially, given a radius of 20, what is the maximum volume that the cone can have. All of the other problems we've covered have involved squares or boxes. I don't even know how to start on this one.
Imagine a disk or radius 20 with a sector of angle $$\displaystyle \theta$$ cut out. The circumference of the entire disk is $$\displaystyle 40\pi$$ and the circumference of the part left after the sector is removed is $$\displaystyle 40(\pi- \theta)$$. A circle of circumference of $$\displaystyle 40(\pi- \theta)$$ has radius $$\displaystyle 20\frac{\pi- \theta}{\pi}$$- that is the radius of the base of the cone.

Look at a cross section of the cone, you should see a right triangle: axis height, base radius, and slant height form a right triangle with slant height as
hypotenuse. The slant height is 20 and the base radius is $$\displaystyle 20\frac{\pi- \theta}{\pi}$$ so, by the Pythagorean theorem, the axis height is $$\displaystyle \sqrt{400- \frac{400(\pi-\theta)^2}{\pi^2}= 20\sqrt{\frac{\pi^2- (\pi- \theta)^2}{\pi^2}$$$$\displaystyle = \frac{20}{\pi}\sqrt{2\pi\theta- \theta^2}$$.

Use those in the formula for volume of a cone: $$\displaystyle V= \frac{1}{3}\pi r^2h$$.

I would really appreciate any help. Between being sick and working, I've having a hard time keeping up, especially when our book has no examples of how to do these kinds of problems.