Relative Minimums and Maximums — Question 10

Question 10 – Solution

Strategy. Solve the marginal-profit equations, then verify the Hessian conclusion by expressing every nearby change as a negative quadratic form.

Step 1: Critical production levels $$P_x=80-4x-y,\qquad P_y=60-x-2y.$$ The equations $4x+y=80$ and $x+2y=60$ yield $$\boxed{x=\frac{100}{7},\qquad y=\frac{160}{7}}.$$

Step 2: Hessian test $$P_{xx}=-4,\qquad P_{yy}=-2,\qquad P_{xy}=-1.$$ Thus $$D=(-4)(-2)-(-1)^2=7>0.$$ Because $P_{xx}<0$, the critical point is a strict relative maximum. Direct substitution gives $$\boxed{P\left(\frac{100}{7},\frac{160}{7}\right)=\frac{8800}{7}}.$$

Step 3: Translated verification Let $h=x-100/7$ and $k=y-160/7$. Since $P$ is quadratic and its linear terms vanish at the critical point, $$P(x,y)-\frac{8800}{7}=-2h^2-hk-k^2.$$ Complete the square: $$-2h^2-hk-k^2 =-2\left(h+\frac{k}{4}\right)^2-\frac 78k^2<0$$ for every $(h,k)\ne(0,0)$. This independently verifies the strict relative maximum.