Answer
a. $y=180-x$
b. $P=x(180-x)$
c. Domain of P is $[0,180]$.
d. $\frac{dP}{dx}=180-2x$ and $x=90$ when $\frac{dP}{dx}=0$.
e. $P(90)=8100$ ; $P(0)=P(180)=0$
f. Maximum $P$ value $= 8100 $ at $x=90; y=90$
Work Step by Step
a. $x+y=180 \implies y=180-x$.
b. substituting in $P$ we get $P=x(180-x)$.
c. Given that both $x$ and $y$ are non-negative so $x\ge0, y\ge0$. Since $x\ge0$, $y\le 180$. Therefore $y=f(x) \in [0,180]$.
d. Differentiating $P$ we get $\frac{dP}{dx}=180-2x$.
$\frac{dP}{dx}=0 \implies x=90$.
e. Checking $P$ values at endpoints $x=0$ and $x=180$ we get $P(0)=P(180)=0$. At the critical point, $P(90)=8100$
f. From step e it is clear that $P$ is maximum at $x=y=90$ and $P(90)=8100$.
