Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 7

$4000$

We are given that $P=xyz~~~(a); \\ x+y=30~~~(b)\\ y+z=30~~~(c)$ or, $x=30-y~~~(d)\\ z=30-y ~~~(e)$ So, $P=(30-y)y(30-y)=y^3-60y^2+900y$ For maximum, we must have $\dfrac{dP}{dx}=0$ or, $3y^2-120y+900=0 \\ y^2-40y+300=0 \\ y^2-30y-10y+300=0\\(y-10)(y-30)=0\\ y=10; 30$ Because $x,y,z \geq 0$, so we can have $y=10$ Now, equations $d$ and $e$ gives: $x=20$ and $z=20$ Thus, our equation (a) becomes: $P_{max}=(20)(10)(20)=4000$