linear optimization: A company has 80 m of steel and....

[Clifford]

A company has 80 m of steel and 200 m of brick. A dog house roof requires 3 m of steel and 6 m of brick. A shed requires 6 m of steel and 12 m of brick. If the profit on a dog house is $100 and the profit on a shed is $50, how many dog houses and sheds need to be build to maximise the profit?

let x be number of dog houses
let y be number of sheds

to maximize profit would be:
P = 100x + 50y

let b be amount of brick
let s be amount of steel
b <= 200
s <= 80

x = 3s + 6b
y = 6s + 12b

I am confused about how I would be able to graph these in order to get the points to test to see about the maximum profit.

I think I have too many variables to be able to graph and I don't know how to eliminate some. Is there another way to approach this only in terms of x and y to make it easier? Could somebody give me a hand?

 

[Denis]

Check the numbers in your problem; don't make much sense:
as it is, build 26 dog houses (no sheds!) for max profit of $2600.