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

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- Dec 14th 2007, 07:33 AMneoseeker191Algebra 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 AMwingless
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 AMSoroban
Hello, neoseeker191!

Quote:

Maximize the function: .$\displaystyle P\:=\:2x+3y$

with the following constraints: .$\displaystyle x \:\geq 0,\quad 0 \leq y \leq 1, \quad 3x+y \:\leq \:3$

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 $\displaystyle P.$