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10-18-2011   #2
MAS1

Join Date: Dec 2008
Posts: 249

Quote:
I'm not sure that you need to use calculus to solve these problems. They looked more like linear programming problems to me, so I used linear programming to solve problem 1.

d = development
a = art
s = design
p = production

Console (c= number of consoles)
----------------
d: 10920/520 = 21 people
a: 13000/520 = 25 people
s: 3120/520 = 6 people
p: 2080/520 = 4 people

Handheld (h = number of handhelds)
----------------------
d: 7280/520 = 14 people
a: 2600/520 = 5 people
s: 9360/520 = 18 people
p: 2600/520 = 5 people

So our limits are given by:

Eq. 1: 21c + 14h <= 238
Eq. 2: 25c + 5h <= 225
Eq. 3: 6c + 18h <= 180
Eq. 4: 4c + 5h <= 57

Since each of these are linear equations I assumed c was the x-axis and h was the y-axis.

For each equation I found the x and y intercepts by setting c equal to 0 and solving for h, then setting h equal to 0 and solving for c.

c | h
-----------------
Eq. 1: 0 | 17
Eq. 1: 34/3 | 0
Eq. 2: 0 | 45
Eq. 2: 9 | 0
Eq. 3: 0 | 10
Eq. 3: 30 | 0
Eq. 4: 0 | 11.4
Eq. 4: 14.25 | 0

I used the intercepts to find the slope of the line for each equation.

m1 = (17 - 0)/(0 - 34/3) = -1.5
m2 = (45 - 0)/(0 - 9) = -5
m3 = (10 - 0)/(0 - 30) = -1/3
m4 = (11.4 - 0)/(0 - 14.25) = -0.8

Rewriting the equations gives:
Eq. 1: h = -1.5c + 17
Eq. 2: h = -5c + 45
Eq. 3: h = -c/3 + 10
Eq. 4: h = -0.8c + 11.4

I then graphed them to see where they intersected each other and the x and y axes. There were 4 points which are easy to see if you graph them.

Point1: (0,10)
Point2: (3,9)
Point3: (8,5)
Point4: (9,0)

Three of the lines (1, 2, and 4) intersect at (8,5).

Then plug in the values for c and h into the equation

Profit = 1.8c + 1h

And pick the largest to find the maximum profit.

Profit1 = 1.8(0) + 10 = 10 million
Profit2 = 1.8(3) + 9 = 14.4 million
Profit3 = 1.8(8) + 5 = 19.4 million
Profit4 = 1.8(9) + 0 = 16.2 million

So the max profit occurs when 8 consoles and 5 handhelds are produced