# 2 Optimization problems!

1. Jul 3, 2011

### QuarkCharmer

1. The problem statement, all variables and given/known data
#56.) Someone makes necklaces and sells them for 10 dollars each. His average sales were 20 per day. When he raises the price to 11 dollars per day, the average sales drops 2.
a.)Find the demand function, assuming it is linear.
b.)If the material to make each necklace costs 6 dollars, what should the selling price be to maximize profit?

#14.)A rectangular container with an open top is to have a volume of 10m^3. The length of it's base is twice the width. Materials for the base cost 10$per square meter. Material for the sides cost 6$ per meter. Find the cost of materials for the cheapest such container.

2. Relevant equations

3. The attempt at a solution
For #56 I have solved part a.) I found the demand function to be
$$y=-2x$$
For part b. I am a bit confused.
It seems to me like I should just let x = (x-6), and then maximize that function? Is that the correct idea?

#14.)For this one I made a box and labeled it's dimensions:
$$h(w)(2w)$$
Then I got thinking about it, and I don't know how to set this one up at all.
I worked out the following:
$$LWH = 10$$
$$L = 2W$$
$$2W^{2}H = 10$$

Area for the sides is:
$$2(2WH)+2(HW)$$
Area for the base is:
$$2W^{2}$$

I just don't know how to relate all of these together? I think it's a similar style to the other problem I posted here, which is why I grouped them together. What is the next step to relate these functions to their appropriate costs so that I can maximize? For the container one, I thought I would just have to find the minimum surface area of the box that still had the volume of 10, but the two different costs ruined that idea for me.

2. Jul 3, 2011

### gb7nash

Part a) is wrong. Do a sanity check on this. If you sell for 0 dollars, by your equation, you sell 0 necklaces.

I also can't tell if your x represents demand or quantity. Usually, for profit problems, you'll want the independent variable to be quantity and the dependent variable to be price. Just so there's no ambiguity as to what you're dealing with, you might want to replace your equation with p's and q's.

Now looking at part a), try to find the equation of the line that fits through the two points: (q1,p1) = (20,10) and (q2,p2) = (18,11)

3. Jul 3, 2011

### QuarkCharmer

I see what you are saying, I made a mistake there.
If two points on the demand function are (10,20) and (11,18)
then
$$\frac{\Delta Q}{\Delta P} =\frac{18-20}{11-10} = -2$$
$$Q = -2P + k$$
$$18 = -2(11) + k$$
$$18 = -22 + k$$
$$18 + 22 = k$$
$$k = 40$$

So the demand function for part a. of that problem is given by:
$$Q = -2P + 40$$

Now, it costs him 6 dollars per necklace, what should his selling price be to maximize profits? I'm still not sure how to relate his cost of 6 per item made to the above formula.

4. Jul 3, 2011

### QuarkCharmer

If Q is how many he is selling, then wouldn't Q(P-6) give his profits?
and so I could write:
$$F(P)=(P-6)(-2P+40)$$
Which would be a polynomial like:
$$F(P)=-2P^{2} + 52P - 240$$
which is opening downwards, so there will be a max!
$$\frac{d}{dp}-2P^{2} + 52P - 240 = -4P+52$$
$$F'(P)= -4P+52 = 0$$
$$-4P = -52$$
$$P = \frac{52}{4}$$
$$P = 13$$

That seems reasonable for that question to me? Or am I mistaken again?

5. Jul 3, 2011

### QuarkCharmer

I guess it's time to tackle the other problem (thanks for the help on the first one):

#14.)A rectangular container with an open top is to have a volume of 10m^3. The length of it's base is twice the width. Materials for the base cost 10$per square meter. Material for the sides cost 6$ per meter. Find the cost of materials for the cheapest such container.

I'm trying to figure out how to set this one up, when I come to something usable I will post here.