Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 8

$256$

We are given that $P=xyz~~~(a); \\ x+z=12~~~(b)\\ y+z=12~~~(c)$ or, $x=12-z~~~(d)\\ y=12-z ~~~(e)$ So, $P=(12-z)(12-z)z=z^3-24z^2+144z$ For maximum, we must have $\dfrac{dP}{dz}=0$ or, $3z^2-48z+144=0 \\ z^2-16z+48=0 \\ z^2-12z-4z+48=0\\(z-4)(z-12)=0\\ z=4,12$ Now, equations $d$ and $e$ gives: $x=0$ and $y=0$ Thus, our equation (a) becomes: $P_{max}=(8)(8)(4)=256$