1. ## 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

2. Let's write the conditions.
$x \geq 0$
$0 \leq y \leq 1$
$3x + y \leq 3$
And we want to maximize: $P = 2x + 3y$

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

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.

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

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

The greater y will make our P bigger. But we have a condition for y, $0 \leq y \leq 1$. The greatest y value we can use is 1. For y=1, x is 2/3. So P is $\frac {13}{3}$.

3. Hello, neoseeker191!

Maximize the function: . $P\:=\:2x+3y$
with the following constraints: . $x \:\geq 0,\quad 0 \leq y \leq 1, \quad 3x+y \:\leq \:3$
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: . $(0,\,0)\,;(1,\,0),\;(0,\,1),\;\left(\frac{2}{3},\, 1\right)$

Now see which one produces maximum $P.$