Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 5

$20$

We are given that $F=x^2+y^2~~~(a)$ and $x+2y=10~~~(b)$ or, $x=10-2y~~~(c)$ So, $F=(10-2y)^2+y^2=5y^2-40y+100$ For minimum, we must have $\dfrac{dF}{dx}=0$ or, $10y-40=0 \\ 10y=40 \\ y=4$ Now, equation $c$ becomes: $x=10-(2)(4)=2$ Thus, our equation (a) becomes: $F_{min}=(2)^2+(4)^2=4+16=20$