Question

Maximize: z=6x+7y

Answer 1

Answer:

So the (x,y) that gives us the maximum z=40 is (40/13 , 40/13).

The problem given:

Maximize: $z=6x+7y$ with:

$5x+8y\le 40$

$12x+y\le 40$

$x\ge 0$

$y\ge 0$

Step-by-step explanation:

I like to solve for y when you have 2 variables in an inequality to tell where to shade.

$5x+8y\le 40$

Subtract 5x on both sides:

$8y \le -5x+40$

Divide both sides by 8:

$y \le \frac{-5}{8}x+\frac{40}{8}$

Simplify:

$y \le \frac{-5}{8}x+5$

The y-intercept is 5 and the slope is -5/8.

The line will be solid due to the =.

We will shade below the line since it says y is less than.

$12x+y\le 40$

Subtract 12x on both sides:

$y\le -12x+40$

The y-intercept is 40 and the slope is -12/1.

The line will be solid due to the =.

We will shade below the line since it says y is less than.

Since the y-intercept is so high, I'm also going to find the x-intercept so I can ignore the slope for now.

The x-intercept of $12x+y=40$ is needed.

The x-intercept can be found be setting y=0 and solving for x:

$12x+0=40$

$12x=40$

Divide both sides by 12:

$x=\frac{40}{12}$

Reduce:

$x=\frac{10}{3}$.

Now for the other two inequality.

$x=0$ is a vertical line; in this case the vertical line is the y-axis.

$y=0$ is a horizontal line; in this case the horizontal line is x-axis.

So for $x\ge$ you will shade to the right of x=0.

For $y\ge$ you shade above y=0.

We are going to include line x=0 and y=0 by leaving them solid.  We are doing this to the equal sign.

Now let's see what we have on the graph.

The intersection of the shading is where you see all the different colors I use.

This is the part where we want to look at.

Let's identify the corners.

One is the purple line's y-intercept which was (0,5).

One is the origin.

The x-intercept of $12x+y=40$ is also included which is at (10/3,0).

Where the green line and purple line cross is another.

We can find the intersection by solving the system:

12x+y=40

5x+8y=40

I'm going to do this by substitution.

Solve the top for y:

12x+y=40

Subtract 12x on both sides:

y=-12x+40

Now we can plug this into the other equation:

5x+8(-12x+40)=40

Distribute:

5x-96x+320=40

Combine like terms:

-91x+320=40

Subtract 320 on both sides:

-91x=40-320

Simplify:

-91x=-280

Divide both sides by -91:

x=-280/-91

Reduce by dividing top and bottom by -7:

x=40/13

We can find the y-coordinate by using y=-12x+40 along with x=40/13.

y=-12(40/13)+40

Put into calculator:

y=40/13

So they cross at (40/13,40/13).

So let's summarize what we have for the corner points:

(40/13 , 40/13)

(0,0)

(10/3,0)

(0,5)

Let's see where the maximum happens for z.

z=6(40/13)+7(40/13)=40

z=6(0)+7(0)=0

z=6(10/3)+7(0)=20

z=6(0)+7(5)=35

The highest number we got was 40.

So the (x,y) that gives us the maximum z=40 is (40/13 , 40/13).