# Thread: Least cost calculation, plz help!

1. ## Least cost calculation, plz help!

Need immediate help on this question. Any help is appreciated!

Cost formulas for objects A and B

A) 3A^2 + 100A+400
B) 6B^2+150B+600
A+B = 500

^2 means to square

What is the combination of A and B that would yield the lowest cost? I don't need the answer. Just the formula if there is one or the steps to take to get to the answer. Thanks a lot

2. Originally Posted by Remeth
Need immediate help on this question. Any help is appreciated!

Cost formulas for objects A and B

A) 3A^2 + 100A+400
B) 6B^2+150B+600
A+B = 500

^2 means to square

What is the combination of A and B that would yield the lowest cost? I don't need the answer. Just the formula if there is one or the steps to take to get to the answer. Thanks a lot
Total cost:

T = 3A^2 + 100A+400 + 6B^2+150B+600

but B=500-A, so substituting this in the formula for T we have:

T = 9 A^2 - 6050 A + 1576000.

To find the lowest cost we differentiate T with respect to A and set the
derivative equal to zero and solve for A.

d/dA T = 6 A - 6050,

so for the minimum: A=6050/6, and B=500 - 6050/6. This gives the point
of global minimum for T (since T is a quadratic in A opening upwards this
stationary point is of necessity an global minimum rather than any other
king of stationary point).

Now I assume that we have to have A,B >=0, so this minimum is not
in the feasible region for this problem, so with these additional constraints
we know that the minimum lies on the boundard of the feasible region.

In this case the feasible region is the line segment from (0,500) to (500,0)
in the AB plane, so the boundary of the feasible region is the pair of points
(500,0) anf (0,500).

That is either A=0 or B=0, When B=0 T=801000, when A=0 T =2026000, so
the minimum occurs when A=500 and B=0, and the total cost is 801000.

RonL

3. Originally Posted by CaptainBlack
Total cost:

T = 3A^2 + 100A+400 + 6B^2+150B+600

but B=500-A, so substituting this in the formula for T we have:

T = 9 A^2 - 6050 A + 1576000.

To find the lowest cost we differentiate T with respect to A and set the
derivative equal to zero and solve for A.

d/dA T = 6 A - 6050,

so for the minimum: A=6050/6, and B=500 - 6050/6. This gives the point
of global minimum for T (since T is a quadratic in A opening upwards this
stationary point is of necessity an global minimum rather than any other
king of stationary point).

Now I assume that we have to have A,B >=0, so this minimum is not
in the feasible region for this problem, so with these additional constraints
we know that the minimum lies on the boundard of the feasible region.

In this case the feasible region is the line segment from (0,500) to (500,0)
in the AB plane, so the boundary of the feasible region is the pair of points
(500,0) anf (0,500).

That is either A=0 or B=0, When B=0 T=801000, when A=0 T =2026000, so
the minimum occurs when A=500 and B=0, and the total cost is 801000.

RonL
I was thinking of doing something along these same lines, but I wasn't sure if he knows calculus. Doing this the algibraic way would involve completing the square, which would be nasty in this case.

4. Originally Posted by ecMathGeek
I was thinking of doing something along these same lines, but I wasn't sure if he knows calculus. Doing this the algibraic way would involve completing the square, which would be nasty in this case.
It had crossed my mind that maybe a non calculus minimisation of the
quadratic might be appropriate, but then if the additional constraint that
A and B>=0 is correct, then this is unlikely to be a precalculus problem.

But meybe I'm wrong!

RonL