quadratic

[stuck_in_mud]

hi, this question is about quadratic programming, and im trying to formulate this problem, but having no luck with it,

the problem is:
A company produces 2 types of products A and B, it costs the company £40 to make product a and £70 to make product B. from research it is suggested that if the seeling price of A and B are set as c and D then they will sell x of A and Y of B, given by the realtionship:
c = £220 - 3x and d = £250-2y.

How would i formulate this problem to give me the answer as:

max: -3x^2-2y^2+180x+180y.

Please help

[hierarch]

Here is my answer.

The seeling price is not less than zero, and the selling price should be not less than the sum of the cost and the seeling price. Then we have
c = 220 - 3x >= 0 (1)
x >= 40+(220-3x) (2)
From equation (1) and (2), the range of x is obtained:
65 <= x <= 220/3 (3)
similarly, we have
d = 250-2y >= 0 (4)
y >= 70+(250-2y) (5)
then, the range of y is obtained:
320/3 <= y <= 125 (6)

and then, the max.
-3x^2-2y^2+180x+180y
=-3(x^2-60x)-2(y^2-90y)
=-3(x-30)^2-2(y-45)^2+3*30^2+2*45^2
<= -3*(65-30)^2-2*(320/3-45)^2+3*30^2+2*45^2 = -4530-5/9

[wolfson]

If they sell x of product A at profit 180-3x each (having subtracted off the £40), how do you find the total profit for product A?
Do the same for product B and you should have your answer.

If that doesn't help, think how much profit you'd make if you sold 5 of A at a profit of £3 each!

[dryan]

When you say "seeling", does that mean ceiling?

You need to make an equation for profit, and find when the derivitive (with respect to pounds) equals zero. This will give you relative maximums and minimums for the equation. (Check endpoints for abs max and min to satisfy the Endpoint Extrema Thrm)