# Algebra Maximizing Function

• Dec 14th 2007, 07:33 AM
neoseeker191
Algebra Maximizing Function
Maximize the function P=2x+3y

with the following constraints:

X is greater than or equal to 0
0 is less than or equal to y is less than or equal to 1
3x+y is less than or equal to 3
• Dec 14th 2007, 09:40 AM
wingless
Let's write the conditions.
$\displaystyle x \geq 0$
$\displaystyle 0 \leq y \leq 1$
$\displaystyle 3x + y \leq 3$
And we want to maximize: $\displaystyle P = 2x + 3y$

We have to maximize both x and y to get a maximum 2x + 3y.
As $\displaystyle 3x + y \leq 3$ and we want to maximize x and y, making it less than 3 won't make x and y maximum. So, $\displaystyle 3x + y = 3$

x can be any positive integer but y is between (or equal to) 0 and 1. This means finding y is better cuz when we use y, we don't have to think of x, as it can be any positive integer.

$\displaystyle 3x + y = 3$
$\displaystyle x = \frac {3-y}{3}$

Then, $\displaystyle P = 2x + 3y$
$\displaystyle P = 2\frac {3-y}{3} + 3y$
$\displaystyle P = 2 + \frac {7y}{3}$

The greater y will make our P bigger. But we have a condition for y, $\displaystyle 0 \leq y \leq 1$. The greatest y value we can use is 1. For y=1, x is 2/3. So P is $\displaystyle \frac {13}{3}$.
• Dec 14th 2007, 11:04 AM
Soroban
Hello, neoseeker191!

Quote:

Maximize the function: .$\displaystyle P\:=\:2x+3y$
with the following constraints: .$\displaystyle x \:\geq 0,\quad 0 \leq y \leq 1, \quad 3x+y \:\leq \:3$

This is a Linear Programming problem.
You should have been shown the approach by now . . .

Graph the inequalities; shade the appropriate regions.
Determine the vertices of the resulting polygon.
Test the vertices (coordinates) in the given function.

The graph should look like this:
Code:

        |       3 |         |\         | \         |  \         |  \         |    \       1 o - - o - -         |::::::\         |:::::::\         |::::::::\       - o - - - - o - -         |        1\

Determine the four vertices.
You should get: .$\displaystyle (0,\,0)\,;(1,\,0),\;(0,\,1),\;\left(\frac{2}{3},\, 1\right)$

Now see which one produces maximum $\displaystyle P.$