Basic algebra/math question.

Factual Questions
1

Say you have this scenario:

Object A costs $10
Object B costs $100

X= A^3 +B^2

What combination of the smallest number of A and B would cost the least? The total number of objects must be at least 30.

This is probably a bad example of what I’m trying to ask here but I need to take a Radar Design problem and figure out the best combination of peak power and antenna area at a cost of $2.20/W and $1400/m^2 respectively while meeting functionality requirements as well.

Is iteration the only way to do this problem? I feel like I am forgetting some basic concept to solve this.

Anyone?

Thanks!

2

I don’t quite understand. You have the following implicit relationships:

C (Cost) = 10A + 100B
N (Number) = A + B

You want to minimize the value of C while keeping N above 30. So far, so good.

What I don’t understand is how the X equation relates. Is X a constant, and if so, what is its value? :slight_smile:

3

for a typical max/min problem I was going to say you need Linear Programming but I’m not sure it’ll work with the problem you laid out because of the exponents.

I also found a section on Nonlinear Programming but niether of these is what I’d call a simple solution.

Hopefully one of our smarter members will be along soon…

But it looks like chrisk may be on to something.

4

I’m going to answer the question that I think the OP is asking. It may or may not relate to the radar design problem that the OP has not asked.

I’m assuming here that exactly 30 objects is acceptable. Otherwise, there’s no unique solution. We’re given that A + B = 30, A > 0, B > 0, X = A[sup]3[/sup] + B[sup]2[/sup], and C = 10A + 100B. We want to maximize X and minimize C–that’s not necessarily possible, so we’ll try maximizing R = X/C.

R = (A[sup]3[/sup] + B[sup]2[/sup])/(10A + 100B). A + B = 30, so B = 30 - A, and R = (A[sup]3[/sup] + (30 - A)[sup]2[/sup])/(10A + 100(30 - A)). The only stationary point of R(A) is a minimum, so the maximum occurs at A = 0 or A = 30. A quick calculation shows that R(30) > R(0), so that’s your maximum.

5

By my calculations, costs are minimized at

A = 11, B = 19

Total cost = $4,941,000

Based on the following:

Cost (X) = (10 A)[sup]3[/sup] + (100 B)[sup]2[/sup]
= 1000 A[sup]3[/sup] + 10000 B[sup]2[/sup]

Units = 30 = A+B

Therefore B = 30 - A

Therefore by subsitution X = 1000 A[sup]3[/sup] + 10000 (30-A)[sup]2[/sup]
= 1000 A[sup]3[/sup] + 10000 (900 - 60A + A[sup]2[/sup])
= 1000 A[sup]3[/sup] + 9000000 - 600000A + 10000 A[sup]2[/sup]

Now the Calculus starts…

X dx = 3000 A[sup]2[/sup] + 20000 A - 600000

And after applying the mighty quadratic formula…

X dx = (11.20 - A)(17.86 + A)

This suggest the optimal value for A is 11 or 12 (since it has to be a whole positive number, I assume, so B would be 19 or 18). Plugging them into the original formula:

X = (10 x 11)[sup]3[/sup] + (100 x 19)[sup]2[/sup]
= $4,941,000

X = (10 x 12)[sup]3[/sup] + (100 x 18)[sup]2[/sup]
= $4,968,000
Go with A = 11, B = 19
And that’s assuming I’ve understood the original question. Don’t get on my case if my notation or terminology are wrong - it’s been a while since Cal 2.