# Thread: Maximizing Profit question

1. ## Maximizing Profit question

This one is giving me some trouble. Can you help me find the solution?

I will be paid 5$for every every book I bring to a man. I will be paid an extra$1 per book per box if I bring them in a box.

It will take me 300 minutes to find 1000 books. It will take me 500 minutes to find 1 box.

I have a given number of minutes to return. What is the best strategy to take in order to earn the most?

What I can figure out so far:

let x be the number of books and y be the number of boxes.

Profit = x * ($5 +$1 * y)
Time = 300(x/1000) + 500y

How do I put the two of these together to figure out the max?

2. My attempt at lagrange has gotten me no where:

f(x,y) = x * (5+y)
g(x,y) = 300(x/1000) + 500y

F(x,y) = f(x,y) - lambda(g(x,y) - k)
F(x, y, lambda) = x * (5 + y) - lambda(300/1000x + 500y - k)
F(x, y, lambda) = 5x + xy - 3/10 lambda x - 500 lambda y + lambda k

Fx = 5 + y - 3/10 lambda - 500y = 0
Fy = x - 500 lambda = 0
Flambda = -3/10x - 500y + k = 0

3. Are the minutes you have to do this not given in the example?

4. Originally Posted by Ond
Are the minutes you have to do this not given in the example?
They are but I'd like to solve for the general case of k. K= 65000

5. Don't know if this is what you're looking for, but here it goes...

We have the following:
1) $\displaystyle F_x=5+y-0,3\lambda-500y=0$
2) $\displaystyle F_y=x-500\lambda=0$
3) $\displaystyle F_\lambda=-0,3x-500y+k=0$

From 2) we have:
$\displaystyle \lambda=x/500$

This into 1) and solve for x gives us:
$\displaystyle 5+y=0,3x/500+500y$
$\displaystyle x=8.333,33+1.667,67y$

We then stick this expression in equation 3) above and solve for y:
$\displaystyle k=0,3(8.333,33+1.667,67y)+500y$
$\displaystyle k=2.500+500y+500y$
$\displaystyle 1.000y=k-2.500$
$\displaystyle y=k/1.000-2,5$

Above where it says "This into 1)..." you just solve for y instead of x and then do the same steps. That will then yield x in terms of k.

Hopefully my calculations are right here, and hopefully this is what you were looking for?