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We are given the following linear programming problem:
Maximize

P = 5 x + 2 y

subject to the following constraints:

x + 2 y \leq 10

5 x + 3 y \leq 30

x \geq 0, y \geq 0

To solve this problem, we can start by graphing the feasible region defined by the constraints.
We begin with the constraint

x + 2 y \leq 10

, which we can rewrite as

y \leq \frac{10}{2} - \frac{1}{2} x

, or

y \leq 5 - \frac{1}{2} x

. This constraint represents a line with a slope of

- \frac{1}{2}

and a

y

-intercept of

5

. To graph this line, we can plot the

y

-intercept and then use the slope to find additional points.
Next, we graph the constraint

5 x + 3 y \leq 30

, which we can rewrite as

y \leq \frac{30}{3} - \frac{5}{3} x

, or

y \leq 10 - \frac{5}{3} x

. This constraint represents a line with a slope of

- \frac{5}{3}

and a

y

-intercept of

10

.
Finally, we add the non-negativity constraints

x \geq 0

and

y \geq 0

by shading in the region of the graph that lies in the first quadrant.
The feasible region is shown below:

\includegraphics [w i d t h = 8cm] g r a p h . p n g

To find the optimal solution, we need to evaluate the objective function

P = 5 x + 2 y

at each of the corner points of the feasible region. These corner points are the intersections of the lines that define the constraints.
We can solve for these corner points by solving the system of equations:

{\begin{matrix} x + 2 y = 10 \\ 5 x + 3 y = 30 \end{matrix}

Solving this system, we find that the intersection point is

(x, y) = (4, 3)

.
Evaluating the objective function at each corner point, we have:

P (0, 0) = 0

P (0, 5) = 10

P (4, 3) = 26

P (6, 0) = 30

Therefore, the optimal solution is

P = 26

, which occurs at the corner point

(4, 3)

.
Thus, the maximum value of

P

is

26

, and this is achieved when

x = 4

and

y = 3

.

Max: P=5x+2y. S.T. x+2y<=10; 5x+3y <= 30, x>=0, y>=0

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