Question 1

2:45 But what are the odds that "12x^2 - 200x + 600" just happens to be in the correct format to use it in the quadratic formula? What if it were: "-200x + 600 = 0" or "12x^3 - 200x + 600 = 0"?

Accepted Answer

Hi Steven. That is an interesting question. Actually it happens to be a quadratic by chance, or more precisely, because the people who invented that exercise wanted the solver to combine his knowledge in calculus and the quadratic formula.

In many other cases it is not so simple to get an exact answer, and you would have to use the "graphing method" Sal explained or make some numerical approximations (either to maximize the function or to set the derivative equal to zero).

Question 2

Why did you use 20-2x to find the domain and not 30-2x?

Accepted Answer

Its because the smallest side is 20", and when x=10" the smallest side is folded and volume=zero.

Question 3

I have an issue with a simular problem. My sheet of metal is given as 40cm x 80cm, and when I follow this method, I end up trying to square root a negative number. Which of course isn't possible, at least without moving into the realm of complex numbers which won't help with my problem.

Accepted Answer

Hi Matthew! I solved this problem for you (mostly for personal practice). I'm not sure how much it will help considering it's 8 years later.

*The following is in Markdown and LaTex. I recommend copying the following content and pasting it into https://stackedit.io/app for easier reading*

```

## Problem
We have a sheet of metal that is $40 	ext{ cm} 	imes 80 	ext{ cm}$ and we want to fold it such that we get the most volume out of it.

Given that, this is what our equation looks like:

$$\begin{align*}
V(x) &= x(40-2x)(80-2x)\
&= x(3200-80x-160x-4x^{2})\
&= x(4x^{2}-240x+3200)\
&= 4x^{3}-240x^{2}+3200x
\end{align*}$$

Derive the above to get:
$$\begin{align*}
V'(x) &= 12x^{2}-480x+3200\
&= 4(3x^{2}-120x+800)
\end{align*}$$

We need to use the quadratic formula to figure out when $V'(x)=0$. Doing so will help us figure out where the absolute maximum and minimum are. Luckily, the highest degree in the polynomial is $x^3$ (which means there is only one maximum and minimum).

$$\begin{align*}
x &= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\
&= \frac{120 \pm \sqrt{(-120)^{2}-4 \cdot 3 \cdot 800}}{2(3)}\
&= \frac{120 \pm \sqrt{4800}}{6}\
&\approx 8.453, \cancel{31.547}\
& \boxed{\approx 8.453}
\end{align*}$$

So, you would need to fold $8.453 	ext{ cm}$ inwards.

Note that the $x$ values we found do **not** need to be changed any further because they both make the quadratic equal $0$.

You might also wonder how I already knew $31.547$ wasn't the answer. The answer is simple. If you plug in $31.547$ into $V(x)$, you get:
$$V(31.547) = 31.547 \cdot \Big(40-2(31.547)\Big) \cdot \Big(80-2(31.547)\Big)$$

Without solving any further, it's obvious that $2(31.547) \approx 63$, which would make it one parentheses negative. Meanwhile, all other terms are positive, so it would make $V(x) <0$ too.

So, all we need to do is find $V(8.453)$:
$$\begin{align*}
V(x) &= x(40-2x)(80-2x)\
V(8.453) &= 8.453(40-2\cdot 8.453)(80 - 2 \cdot 8.453)\
&= 8.453(40-16.906)(80 - 16.906)\
&= 8.453(23.094)(63.094)\
&\approx 12,316.805
\end{align*}$$

You could also try verifying that this is the *absolute* maximum by finding $V''(x)$, but it isn't necessary because (as stated before) $x^3$ is the highest degree. That means only one maximum and one minimum (which we found using the quadratic formula from above).

## Conclusion
To conclude, the most efficient way to fold the $40 	ext{ cm} 	imes 80 	ext{ cm}$ is to fold each corner in $8.453 	ext{ cm}$. Doing so gives you $\approx 12,316.805 	ext{ cm}^3$.

Alternatively, you could just go onto Desmos and plug in equation from the beginning to find the maximum value.
```

Question 4

Instead of doing the second derivative test, couldn't you have just plugged in the x values into the original equation?

Accepted Answer

You can do it that way, but it requires more points to determine the concavity.  In most cases, the second derivative test is the easiest way to do it.

Question 5

How does Khan simplify 4^3 - 100x^2 + 600x to 12x^2 - 200x + 600? I am very new to calculus and just came to watch these videos for optimization, so please explain very clearly.

Accepted Answer

Try searching Khan Academy for 'power rule'.

Question 6

So, what units would this problem be answered in? Would it be unitless, or would you say "units cubed?"

Accepted Answer

It would be inches^3 as the original measurements of the cardboard were given in inches

Question 7

In maxima-minima word problems, if something is not given as a constant, and is not mentioned at all, can we take it to be variable?

Accepted Answer

This question is tough to answer because it is not clear exactly what kind of situation you're in.

The word problem could be a general problem, meaning that you should solve the problem for any imaginable number, which gives you an answer, in this case a formula for solving that type of problem.

If it's a specific problem, you can see the previous video on the graphical solution to find out how Sal managed to find an expression for the different lengths:

Basically - if you want to find an exact solution you need to be given enough variables to be able to compute the exact answer in the end.

That does not mean these things are written down for you, it might be implicit in the problem itself, and the problem might be constructed for you to be able to deduce the necessary equation that allows you to find the maximum/minimum of the function

Hope that helps!

Question 8

isnt x also not equal to 15 because of the polynomial (30-2x)?

If x = 15 then (30-2x)=0 and then V(x) will also be equal to 0

Accepted Answer

You are right, but 𝑥 ≠ 15 is already included in 0 ≤ 𝑥 ≤ 10.

Question 9

I'm having problems understanding what I'm looking for. I don't quite fully understand the concept of optimization in total. I have a test today in math and I'm scared I'm not going to do well. Could you please help me understand this? Thank you :)

Accepted Answer

I don’t know how much help you can get from a brief reply here, but I’ll offer some comments for what they’re worth.

Optimization problems have to do with finding a tipping point. Something is getting better up to a point, and then it starts to get worse. It’s getting bigger, then it starts to get smaller. Or it’s getting smaller, then it starts to get bigger.

We find those tipping points by looking at the derivative, which is the rate at which something is changing. As long as the rate of change is in the “good” direction (which may be up or down, depending on what you’re optimizing), we keep going. The tipping point is where the derivative starts to go in the other direction. It’s the top of a frown-shaped curve or the bottom of a smile-shaped curve. The rate of change at these exact moments is zero, so we hunt for optimization points by finding the derivative and then determining where the derivative is equal to zero (the “critical points”).

You need to be aware that some functions will have two or more critical points, and you have to use other tests to determine which one is optimal in terms of the way the problem was set forth. Some functions have critical points that aren’t turning points (for example, y = x^3 has a critical point at x = 0 but doesn’t turn down). And some teachers want to test your understanding by giving you a problem where there’s only one critical point, which is a true turning point, but because of the way the problem was stated, this is a worst solution instead of an optimal one.

Good luck!

Question 10

how would you find the dimensions of an open top box with a  minimum surface area and  a volume of 32000 cubic cm, I'm guessing partial derivatives but lost other than that

Accepted Answer

In theory you have three variables to work with: height, width and depth. In practice, it's fairly obvious that the box has to be square, so you have only two variables, one for the height of the box and one for the length of a side (width and depth). If you're expected to prove this part, you begin by showing that the rectangle with minimum perimeter for a fixed area is a square.

With the problem reduced to height and length of a side, you can create a formula for the surface area (remembering to include the bottom of the box) and set it equal to 32,000. Differentiate and find the place where the derivative is equal to zero, then confirm that this is a minimum.

Course: Differential Calculus > Unit 5

Optimization: box volume (Part 2)

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