Maximisation and Minimisation in Geometry Question (1 Viewer)

tk8 · Feb 20, 2022

I've attached the whole question, but I only need help with:

Why is the answer to part c) ≈1.06? I got ≈0.15
Why is the answer to part d) ≈11.79? I got ≈4.22

Any help with clarifying my errors will be greatly appreciated!

Screen Shot 2022-02-20 at 6.40.04 pm.png

5uckerberg · Feb 20, 2022

@tk8 Here is the solution. I want you to check your work to see where you tripped yourself in the path of solving these two parts.

Part c

To start off $V=10\pi{r^{2}}-2\pi^{2}r^{3}$
Then, $\frac{dV}{dr}=20\pi{r}-6\pi^{2}r^{2}$
Note this means the rate of volume divided by the rate of radius.
So if we want to find the maximum or minimum then the rate has to be resting or in this case 0.
Using this knowledge we have determined that
$20\pi{r}-6\pi^{2}r^{2}=0$
Factorising we can automatically see that
$2\pi{r}\left(10-\mathbf{3}\pi{r}\right)=0$
Note here $r=0 \cup \frac{10}{3\pi}$ .
To confirm do the second derivative $\frac{d^{2}V}{dr^{2}}=20\pi-12\pi^{2}r$
Did thee question say there was a minimum volume? Well, nope it said the maximum volume.
So instead sub in $r=\frac{10}{3\pi}$ and you should get $-20\pi$ which is considerably smaller than 0 so thus, you have confirmed that there is a maximum.
Can you spot where you went wrong in your working

Part d

Since we have proven that $r=\frac{10}{3\pi}$ then put $r=\frac{10}{3\pi}$ into $V=\pi{r^{2}}\left(10-2\pi{r}\right)$
Watch as the magic unfolds
$V=\pi\times{\frac{100}{9\pi^{2}}\left(10-2\pi{\times{\frac{10}{3\pi}}}\right)$
$\frac{100}{9\pi}\left(\frac{10}{3}\right)$
This simply becomes
$\frac{1000}{27\pi}$
and you can do the rest
There must have been an error carried forward. Find it.

j8aab · Feb 20, 2022

have a look at my solutions attached

tk8 · Feb 20, 2022

j8aab said:
have a look at my solutions attached

Thanks so much!

tk8 · Feb 20, 2022

5uckerberg said:
@tk8 Here is the solution. I want you to check your work to see where you tripped yourself in the path of solving these two parts.

Part c

To start off $V=10\pi{r^{2}}-2\pi^{2}r^{3}$
Then, $\frac{dV}{dr}=20\pi{r}-6\pi^{2}r^{2}$
Note this means the rate of volume divided by the rate of radius.
So if we want to find the maximum or minimum then the rate has to be resting or in this case 0.
Using this knowledge we have determined that
$20\pi{r}-6\pi^{2}r^{2}=0$
Factorising we can automatically see that
$2\pi{r}\left(10-\mathbf{3}\pi{r}\right)=0$
Note here $r=0 \cup \frac{10}{3\pi}$ .
To confirm do the second derivative $\frac{d^{2}V}{dr^{2}}=20\pi-12\pi^{2}r$
Did thee question say there was a minimum volume? Well, nope it said the maximum volume.
So instead sub in $r=\frac{10}{3\pi}$ and you should get $-20\pi$ which is considerably smaller than 0 so thus, you have confirmed that there is a maximum.
Can you spot where you went wrong in your working

Part d

Since we have proven that $r=\frac{10}{3\pi}$ then put $r=\frac{10}{3\pi}$ into $V=\pi{r^{2}}\left(10-2\pi{r}\right)$
Watch as the magic unfolds
$V=\pi\times{\frac{100}{9\pi^{2}}\left(10-2\pi{\times{\frac{10}{3\pi}}}\right)$
$\frac{100}{9\pi}\left(\frac{10}{3}\right)$
This simply becomes
$\frac{1000}{27\pi}$
and you can do the rest
There must have been an error carried forward. Find it.

I appreciate the detail, thanks!

tk8 · Feb 21, 2022

Could you guys also please help me with part c) and part d) of the attached question?

I don't know how to prove the maximisation for part c), and I am not attaining the correct ratio for part d).

Thanks

Screen Shot 2022-02-21 at 10.44.01 am.png

5uckerberg · Feb 21, 2022

tk8 said:
Could you guys also please help me with part c) and part d) of the attached question?

I don't know how to prove the maximisation for part c), and I am not attaining the correct ratio for part d).

Thanks

View attachment 35017

To start off we need to know the rate of change in volume per the rate of change in height which is written as $\frac{dV}{dh}=\pi\left(R^{2}-\frac{3h^{2}}{4}\right)$ .

Once the foundation is set we want to find the maximum or minimum and we need to show that by stating the rate of change is zero as you know the rate of change interrupts us from finding the maximum or the minimum.

In this case let $\frac{dV}{dh}=0 \rightarrow \pi\left(R^{2}-\frac{3h^{2}}{4}\right)=0$
In fact $\pi$ is a spectator and as such he can get evicted from the equation.
There, $R^{2}-\frac{3h^{2}}{4}=0\rightarrow R^{2}=\frac{3h^{2}}{4}$
$\times{4}$ will be done on both sides.
$4R^{2}=3h^{2}$
Finsihing it off $h=\frac{2R}{\sqrt{3}}$ or $h=\frac{2}{3}\sqrt{3}R$ if you want to rationalise the denominator. The reason is we are using length so if we square root the solution they stay positive.
Somebody wanted to rationalise the cylinder.

part d)
Sub in $h=\frac{2}{3}\sqrt{3}R$ into $V=\frac{\pi{h}}{4}\left(4R^{2}-h^{2}\right)$
There we will have $V=\frac{\pi}{4}\times{\frac{2R}{\sqrt{3}}\left(4R^{2}-\frac{4R^{2}}{3}\right)$
At this point we will have $V=\frac{\pi{R}}{2\sqrt{3}}\left(\frac{8R^{2}}{3}\right)$
Simplifying further it becomes $V=\frac{4\pi{R^{3}}}{3\sqrt{3}}$

At this point compare with the volume of the sphere so now we have
Volume of sphere : Volume of cylinder
$\frac{4\pi{R^{3}}}{3} : \frac{4\pi{R^{3}}}{3\sqrt{3}}$
At this point you can say the ratio of the volume of sphere to the maximum value of the cylinder is $\sqrt{3} : 1$

Have a good lunch break @tk8

5uckerberg · Feb 21, 2022

To prove part c for the maximum do the second derivative. Note that if the result is negative then you have proven that it’s the maximum.

Maximisation and Minimisation in Geometry Question (1 Viewer)

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