For $x,y$ satisfy $x+y-1=\sqrt{2x-4}+\sqrt{y+1}$, minimize and maximize $$\left(x+y\right)^2-\sqrt{9-x-y}+\frac{1}{\sqrt{x+y}}$$
For $x,y$. Minimize and Maximize $\left(x+y\right)^2-\sqrt{9-x-y}+\frac{1}{\sqrt{x+y}}$
-1
$\begingroup$
maxima-minima
-
0What did you try? – 2017-02-19
-
0haven't ideas.. – 2017-02-19
-
0wait i think it very easy – 2017-02-19
-
1follow my answer and fix for me – 2017-02-19
1 Answers
0
$$S=\left(x+y\right)^2-\sqrt{9-x-y}+\frac{1}{\sqrt{x+y}}$$
We have $$\sqrt{2x-4}+\frac{1}{\sqrt{2}}\sqrt{2(y+1)}\leq\sqrt{3(x+y-1)}$$
From $$x+y-1=\sqrt{2x-4}+\sqrt{y+1}\Rightarrow x+y-1\leq\sqrt{3(x+y-1)}$$ Hence $$1\leq(x+y)\leq4$$
Let $$t=x+y;1\leq t\leq4$$ we have
$$S=t^2-\sqrt{9-t}+\frac{1}{\sqrt{t}}$$
Survey on function $S(t)$ on domain $[1;4]$ (covariates function)
We find $P_{max}=\frac{33-2\sqrt{5}}{2}$ when $x=4;y=0$
$P_{min}=2-2\sqrt{2}$ when $x=2;y=-1$