Review : Systems of Equations — Question 10

Question 10 – Solution

Strategy. Solve the equalities first, then optimize over the remaining feasible interval rather than over an unrestricted algebraic family.

Step 1: Parameterize and restrict. Subtract the first equality from the second to obtain $y+2z=4$. Set $z=t$, giving $$(x,y,z)=(2+t,4-2t,t).$$ Nonnegativity imposes $t\ge 0$, $t\le 2$ and $t\ge-2$. The complete feasible set is therefore $$\boxed{(x,y,z)=(2+t,4-2t,t),\qquad 0\le t\le 2.}$$ Each member satisfies both original totals; every feasible member must arise from this parameterization.

Step 2: Optimize on the feasible interval. The return becomes $P=4(2+t)+2(4-2t)+t=16+t$. It is strictly increasing, so the unique minimum is $\boxed{16}$ at $(2,4,0)$, and the unique maximum is $\boxed{18}$ at $(4,0,2)$. Neither an interior stationary-point calculation nor an integer search is needed; the stated amounts may be fractional.

Step 3: Include the new capacity. The constraint $x\le 3$ becomes $2+t\le 3$, or $t\le 1$. The new feasible interval is $[0,1]$, with unique maximizing allocation $$\boxed{(x,y,z)=(3,2,1),\qquad P=17.}$$ The minimum remains at $t=0$. In the figure, the feasible sets are shown in the $(z,y)$ plane; $x=2+z$ is determined by the equalities.

Step 4: Verify the optimizers and the role of inequalities. The endpoint triples $(2,4,0)$ and $(4,0,2)$ have totals $(6,10)$ and are nonnegative. The new optimizer $(3,2,1)$ also has totals $(6,10)$ and satisfies $x=3$. The former maximizer has $x=4$, so it is no longer allowed. Over the unrestricted equality family, $P=16+t$ has no finite maximum or minimum. The inequalities make the allowed interval bounded and turn this into a well-posed allocation problem.