a.
Let:
M = maximum number of hours that Daniel can work per week
x = Number of hours spend a week parking cars
y = Number of hours spend working at the Haul- Mart store
C(x,y) = Maximum earnings.
b. Daniel works no more than 16 hours a week while attending college, Therefore:
[tex]x+y\leq16[/tex]
He wants to spend at least 2 hours but no more than 10 hours a week parking cars, hence:
[tex]\begin{gathered} x\ge2;x\leq10 \\ or \\ 2\leq x\leq10 \end{gathered}[/tex]
c. He is paid $8 per hour parking cars (x) and $7 per hour working at the Haul-Mart store (y):
[tex]C(x,y)=8x+7y[/tex]
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the maximum number of hours he wants to work in the parking lot is 10, so:
[tex]\begin{gathered} x+y\leq16 \\ \text{If x=10} \\ 10+y\leq16 \\ \text{Solving for y:} \\ y\leq16-10 \\ y\leq6 \end{gathered}[/tex]Replacing those values into the cost function:
[tex]\begin{gathered} C(x,y)=8x+7y \\ C(10,6)=8(10)+7(6)=80+42=122 \end{gathered}[/tex]