Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 6

$12$

We are given that $F=x^2+y^2~~~(a)$ and $xy^2=16~~~(b)$ or, $y^2=\dfrac{16}{x}~~~(c)$ So, $F=x^2+\dfrac{16}{x}$ For minimum, we must have $\dfrac{dF}{dx}=0$ or, $2x-\dfrac{16}{x^2}=0 \\ x^3=8 \\ x=2$ Now, equation $c$ becomes: $y^2=\dfrac{16}{2} \implies y=2\sqrt 2$ Thus, our equation (a) becomes: $F_{min}=(2)^2+(2 \sqrt 2)^2=4+8=12$