How can I solve this problem. I can find absolute maximum and minimum value of equation with given interval. But here, I don't know where should I start. Could you explain step by step?!
[Math] Under condition $2x^2 + y^2= 4$ for real numbers $x, y$, find the maximum and minimum value of $4x + y^2$.
algebra-precalculusdiscriminantmaxima-minimaoptimizationquadratics
Best Answer
Let $4x+y^2=k$.
Thus, $$2x^2-4x+k-4=0,$$ which gives $$4-2(k-4)\geq0$$ or $$k\leq6.$$ The equality occurs for $x=\frac{4}{2\cdot2}=1$ and $y=\sqrt{2},$ which says that $6$ is a maximal value.
In another hand, since $-\sqrt2\leq x\leq\sqrt2,$ we obtain: $$4x+y^2=4x+4-2x^2=2(3-(x-1)^2)\geq2(3-(-\sqrt2-1)^2)=-4\sqrt2.$$ The equality occurs for $x=-\sqrt2$ and $y=0,$ which says that we got a minimal value.