1

I'm faced with this expression:

$$max(xy) : x \leq a, \; y \leq b,\; c = \frac{x}{y}$$ where all variables are greater than or equal to zero.

I know the values for a, b, and c but I need to find the maximum value for x * y and the values for x * y that produce the maximum.

I've started working on this on paper but I'm starting to wonder if it actually can be done. I enjoy deriving things so I'm not looking for an answer yet, would just like to know if I can somehow prove that it can be done.

rtheunissen
  • 103
  • What is $c$? As it stands, it is unbounded, since I can choose $x$ and $y$ arbitrarily large and negative, to make $xy$ as large as I choose. – Daryl Apr 16 '13 at 09:18
  • I'm not very good with formatting so if anyone would like to do that it might help others to understand it. @Daryl, I've edited the question to indicate that the numbers have to be positive integers (0 otherwise). The values for a, b and c are arbitrary but known at the time. – rtheunissen Apr 16 '13 at 09:22

2 Answers2

2

You have $x=cy$, so $xy=cy^2$ is what you're trying to maximise. Noting that this is an increasing function of $y$, you should choose the maximum possible value for $y$ allowed by the constraints. This could be $y=b$ provided $cb \le a$, or $y=\dfrac{a}{c}$ otherwise.

Macavity
  • 46,381
1

Given the clarification, the third constraint gives $x=cy$.

The problem is then $\max\limits_y cy^2$ subject to $cy\leq a,\,y\leq b,\, y\geq 0$.

Choosing $y=\min\left\{b,\dfrac{a}{c}\right\}$ will then satisfy all three constraints with the largest objective function. The corresponding value for $x$ can be calculated.

Daryl
  • 5,598