find the maximum for the profit function, p = 2x + 5y subject to the following constraints. (\begin{cases}2x…

Explanation:

Step1: Find intersection points

Solve the systems of equations:

1. Intersection of $2x + y=80$ and $x = 0$: Substitute $x = 0$ into $2x + y=80$, we get $y = 80$. The point is $(0,80)$.
2. Intersection of $2x + y=80$ and $y = 0$: Substitute $y = 0$ into $2x + y=80$, we get $2x=80$, so $x = 40$. The point is $(40,0)$.
3. Intersection of $x + 2y=80$ and $x = 0$: Substitute $x = 0$ into $x + 2y=80$, we get $2y=80$, so $y = 40$. The point is $(0,40)$.
4. Intersection of $x + 2y=80$ and $y = 0$: Substitute $y = 0$ into $x + 2y=80$, we get $x = 80$. The point is $(80,0)$.
5. Intersection of $2x + y=80$ and $x + 2y=80$: From $2x + y=80$, we have $y=80 - 2x$. Substitute it into $x + 2y=80$: [ \begin{align*} x+2(80 - 2x)&=80\ x + 160-4x&=80\

• 3x&=80 - 160\ -3x&=-80\ x&=\frac{80}{3} \end{align*} ] Then $y=80-2\times\frac{80}{3}=\frac{240 - 160}{3}=\frac{80}{3}$. The point is $(\frac{80}{3},\frac{80}{3})$.

Step2: Evaluate profit function at intersection points

1. At $(0,80)$: $P=2\times0 + 5\times80=400$.
2. At $(40,0)$: $P=2\times40+5\times0 = 80$.
3. At $(0,40)$: $P=2\times0+5\times40 = 200$.
4. At $(80,0)$: $P=2\times80+5\times0=160$.
5. At $(\frac{80}{3},\frac{80}{3})$: $P=2\times\frac{80}{3}+5\times\frac{80}{3}=\frac{160 + 400}{3}=\frac{560}{3}\approx186.67$.

Answer:

$400.00$