I Have no idea where to start, there bonus questions and the teacher asked us to use the internet as our resource
If you guys could show work or explain how to do it that'd be great. I will pay some fg if this gets completed in a timely fashion

Noob

1a. find f'(x) and set it equal to 0 and solve for x. For this function the derivative will be quadratic so you will probably have 2 solutions. These are the answers to part a.

b. put these solutions(from a) on a number line, and test each interval to see where it is increasing or decreasing.

c. if the derivative before a critical point is positive, and negative after the critical point, this is a local maximum; if f'(x) is negative before, and positive after it is a local minimum. It is possible a critical point is neither.

d. take the second derivative and test it at the critical points (from part a). If this results in a negative value, it is a local max. if it is positive, it is a local min.

I'll look at the rest in a while. If you have problems, or want us to check your solutions, post it up.

This post was edited by Azrad on Nov 1 2013 05:25pm

Linear approximations use the derivative to attempt to estimate rough values.

In this case you know sqrt(49) = 7

Assume

f(x) = sqrt(x)

f'(x) = 1/2 x^(-1/2)

using taylor's theorem:
f(x) ~ f(a) + f'(a) * (x-a)

sqrt(50) = f(50) ~ f(49) + f'(49) * (50-49)
= sqrt(49) + 1/2*49^(-1/2) * 1
=7 + 1/28
~7.03

sqrt(50) ~ 7.07 so we're pretty close

This post was edited by nlin on Nov 1 2013 05:37pm

#2:

You are given a volume, so use that equation and solve for one of your variables. I'll pick h:
V = (pi)r^(2)h = 500
h = 500/(pi)r^2

Then use the surface area equation and plug in your h-value you just found:
SA = (pi)r^2 + 2(pi)rh
SA= (pi)r^2 + 2(pi)r(500/(pi)r^2)
SA= (pi)r^2 + 1000/r

Differentiate and set to 0, solve for r:
dSA/dr = 2(pi)r - 1000/r^2 = 0
2(pi)r = 1000/r^2
r^3 = 500/(pi)
r = (500/(pi))^1/3
r ≈ 5.42 cm

Now that you found one of your variables, plug that value into your original equation to find h:
h = 500/(pi)r^2
h = 500/[(pi)(5.42)^2]
h = 5.42cm

To use the Second Derivative Test, you take the derivative of the dSA/dr equation again and then plug in your r:

SA''(r) = 2(pi) + 2000/r^3
SA''(5.42) = 2(pi) + 2000/(5.42)^3 = 18.84 > 0
Since it is positive, then it is a local minimum

We need to know the formula for the surface area of a can. You can either look it up, or figure it out yourself. I'll figure it out myself:
A can contains 2 circles of the same size (the top and bottom), their area is: pi*r^2 (well times 2 since there are two circles).
and a rectangle "rolled" into the uhh tube of the can, area is: w*h

To the surface area of our can is:
A{r,w,h} = 2*pi*r^2 + w*h

now our surface area is a function of 3 variables, which is probably freaking you out. We need to reduce it to 1 variable. Radius looks like a good variable to use, so lets try to get rid of w and h.

first off, if you think about it, the width of the rectangle part of the can (the tube) is the same as the circumference of the lid (circle). So:
w=2*pi*r

So lets rewrite our function:

A{r,h} = 2*pi*r^2+2*pi*r*h

ok now we need to get rid of h:

The volume of the can is 500=pi*r^2*h
solving that for h and we will have h in terms of r, which we can then use to substitute into our surface area function.

h = 500/(pi*r^2)

rewriting surface area again:


A{r} = 2*pi*r^2+2*pi*r*500/(pi*r^2)

cleaning that up a little:

A{r} = 2*pi*r^2 + 1000/r

Now it is just a matter of taking the derivative with respect to r, setting that equal to 0, and solving for r to find the critical point(s). Test these points to find which one (if there is more than one), is a minimum. Then you will know the value for r that minimizes the surface area of the can. You can plug back into the equation h = 500/(pi*r^2) to find the height once you know the radius.

/e
I'm using the same logic as TritonV8, although I got a different answer. Hehe, so one of us made a minor mistake somewhere, I'll try to find it.

/ee
Think I found it:

Ah Triton is modelling the surface area of a "can" with no lid, and I'm modelling for one with a lid. That is the difference.

This post was edited by Azrad on Nov 1 2013 06:17pm

oooh, nice catch. i passed over the closed can part!

What would be the f'(x) for number one? Could you use the power rule?
aka....
8x-3x^2?

correct

the first time I ever did a problem like this, there were more constraints (like a minimum and maximize allowed radius) and I was looking to minimize the cost of the can and the lid/bottom cost was different per unit area than the sides. So i worked the whole thing out myself, felt very proud, got a critical point and submitted it as the answer, AND DIDN'T TEST THE ends of the interval (the min and max allowed radius)..... guess what... my critical point was a maximum, so I submitted the "worst possible answer"...

This post was edited by Azrad on Nov 1 2013 07:16pm