find the maximum for the profit function, p = 15x + 25y subject to the following constraints…

Answer:

$375.00$

Explanation:

Step1: Find corner - points

1. Intersection of $3x + 4y=60$ and $x = 0$:
   • Substitute $x = 0$ into $3x + 4y=60$, we get $4y=60$, so $y = 15$. The point is $(0,15)$.
2. Intersection of $3x + 4y=60$ and $y = 0$:
   • Substitute $y = 0$ into $3x + 4y=60$, we get $3x=60$, so $x = 20$. The point is $(20,0)$.
3. Intersection of $x + 8y=40$ and $x = 0$:
   • Substitute $x = 0$ into $x + 8y=40$, we get $8y=40$, so $y = 5$. The point is $(0,5)$.
4. Intersection of $x + 8y=40$ and $11x + 28y=380$:
   • From $x=40 - 8y$, substitute it into $11x + 28y=380$.
   • $11(40 - 8y)+28y=380$.
   • $440-88y + 28y=380$.
   • $- 60y=380 - 440=-60$, so $y = 1$.
   • Then $x=40-8\times1 = 32$. The point is $(32,1)$.
5. Intersection of $3x + 4y=60$ and $11x + 28y=380$:
   • Multiply $3x + 4y=60$ by $7$ to get $21x+28y = 420$.
   • Subtract $11x + 28y=380$ from $21x+28y = 420$:
     • $(21x+28y)-(11x + 28y)=420 - 380$.
     • $10x=40$, so $x = 4$.
     • Substitute $x = 4$ into $3x + 4y=60$, we have $3\times4+4y=60$, $12 + 4y=60$, $4y=48$, $y = 12$. The point is $(4,12)$.

Step2: Evaluate profit function at corner - points

1. For $(0,15)$: $P=15\times0+25\times15=375$.
2. For $(20,0)$: $P=15\times20+25\times0 = 300$.
3. For $(0,5)$: $P=15\times0+25\times5=125$.
4. For $(32,1)$: $P=15\times32+25\times1=480 + 25=505$.
5. For $(4,12)$: $P=15\times4+25\times12=60 + 300=360$.

The maximum value of $P$ is $375$.