QuestionMaximize $P=3x + 2y$ with constraints: $5x + y \leq 16$ , $2x + 3y \leq 22$ , $x \geq 0$ , $y \geq 0$ .

Studdy Solution

STEP 1

Assumptions1. We are given a linear function $=3x +y$ . . We are given four constraints \begin{align} 5x + y &\leq16, \\ x +3y &\leq22, \\ x &\geq0, \\ y &\geq0. \end{align}
3. We are asked to find the maximum value of the function subject to these constraints.

STEP 2

This is a linear programming problem. The maximum or minimum value of the function (in this case, the maximum value of $$) will occur at a vertex of the feasible region defined by the constraints. The feasible region is the set of all points $(x, y)$ that satisfy all the constraints.

STEP 3

First, we need to find the vertices of the feasible region. These are the points where the constraint lines intersect. We can find these points by solving the system of equations formed by each pair of constraint lines.

STEP 4

Let's start with the first two constraints\begin{align} x + y &=16, \\ 2x +3y &=22. \end{align}

STEP 5

We can solve this system of equations using substitution or elimination method. Let's use the substitution method. First, solve the first equation for $y$ $y =16 -5x.$

STEP 6

Substitute $y =16 -5x$ into the second equation $2x +3(16 -5x) =22.$

STEP 7

implify and solve for $x$ $2x +48 -15x =22,$
$-13x = -26,$
$x =2.$

STEP 8

Substitute $x =2$ back into the first equation to find $y$ $5(2) + y =16,$
$y =6.$
So, one of the vertices of the feasible region is $(2,6)$ .

STEP 9

Next, we need to find the intersection of the constraint lines $5x + y =16$ and $x =$ .

STEP 10

Substitute $x =0$ into the equation $5x + y =16$ to find $y$ $5(0) + y =16,$
$y =16.$
So, another vertex of the feasible region is $(0,16)$ .

STEP 11

Similarly, find the intersection of the constraint lines $x +3y =22$ and $x =0$ .

STEP 12

Substitute $x =0$ into the equation $2x +y =22$ to find $y$ $2(0) +y =22,$
$y = \frac{22}{}.$
So, another vertex of the feasible region is $(0, \frac{22}{})$ .

STEP 13

Finally, find the intersection of the constraint lines $x =0$ and $y =0$ .

STEP 14

This is simply the origin, $(0,0)$ .

STEP 15

So, the vertices of the feasible region are $(2,)$ , $(0,)$ , $(0, \frac{22}{3})$ , and $(0,0)$ .

STEP 16

Now, we need to evaluate the function $=3x +2y$ at each of these vertices to find the maximum value.

STEP 17

Evaluate at $(2,6) $=3(2) +2(6) =6 +12 =.$ $

STEP 18

Evaluate at $(0,16) $=3(0) +2(16) =0 +32 =32.$ $

STEP 19

Evaluate at $(, \frac{22}{3}) $=3() +(\frac{22}{3}) = + \frac{44}{3} = \frac{44}{3}.$ $

STEP 20

Evaluate at $(0,0) $=3(0) +(0) =0.$ $

STEP 21

The maximum value of $$ is the largest of these values, which is $32$.
So, the maximum value of the function $=3x +y$ subject to the given constraints is $32$ .

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