Question

find the maximum value of p = 9x + 8y subject to the following constraints: now, find the coordinates of the corner points.
graph with x-axis, y-axis, a purple triangle, a table with x and y values (0,0; 6,0; 0,7; 3, ?), and the constraints: ( \begin{cases} 8x + 6y leq 48 \\ 7x + 7y leq 49 \\ x geq 0 \\ y geq 0 end{cases} )

Explanation:

Step1: Identify the constraint for x=3

We know the constraints are \(8x + 6y\leq48\) and \(7x + 7y\leq49\). Let's use the second constraint \(7x + 7y = 49\) (since at the corner point, the inequality becomes an equality for one of the constraints; let's check which one is active when \(x = 3\)). Substitute \(x = 3\) into \(7x+7y = 49\).
The equation is \(7(3)+7y=49\).

Step2: Solve for y

First, calculate \(7(3)=21\). So the equation becomes \(21 + 7y=49\). Subtract 21 from both sides: \(7y=49 - 21=28\). Then divide both sides by 7: \(y=\frac{28}{7} = 4\). We can also check with the first constraint \(8x + 6y\) when \(x = 3,y = 4\): \(8(3)+6(4)=24 + 24=48\), which satisfies \(8x + 6y\leq48\) (it's equal, so it's on the boundary). So the corner point when \(x = 3\) has \(y = 4\).
Now, let's find all corner points:

1. Intersection of \(x = 0\) and \(y = 0\): \((0,0)\)
2. Intersection of \(8x + 6y=48\) and \(y = 0\): Solve \(8x=48\Rightarrow x = 6\), so \((6,0)\)
3. Intersection of \(7x + 7y=49\) and \(x = 0\): Solve \(7y=49\Rightarrow y = 7\), so \((0,7)\)
4. Intersection of \(8x + 6y=48\) and \(7x + 7y=49\): We already found when \(x = 3\), \(y = 4\), so \((3,4)\)

Now, to find the maximum of \(P = 9x+8y\):

• At \((0,0)\): \(P=9(0)+8(0)=0\)
• At \((6,0)\): \(P=9(6)+8(0)=54\)
• At \((0,7)\): \(P=9(0)+8(7)=56\)
• At \((3,4)\): \(P=9(3)+8(4)=27 + 32=59\)

Answer:

The corner points are \((0,0)\), \((6,0)\), \((0,7)\), \((3,4)\). The maximum value of \(P\) is \(59\) (achieved at \((3,4)\)), and the missing \(y\) - value when \(x = 3\) is \(4\).