# find the maximum and minimum value of the function $x^3+y^3-3x-12y+10$

find the maximum and minimum value of the function $x^3+y^3-3x-12y+10$? here the question is of $2$ variables and i am not able to solve that ,i know how to solve maximum and minimum question of 1 variable,so please help me to solve this

For positive variables by AM-GM we obtain: $$x^3+y^3-3x-12y+10=x^3+1+1+y^3+8+8-3x-12y-8\geq$$ $$\geq3\sqrt[3]{x^3\cdot1\cdot1}+3\sqrt[3]{y^3\cdot8\cdot8}-3x-12y-6=-8.$$ The equality occurs for $x=1$ and $y=2$, which says that $$\min_{x>0,y>0}(x^3+y^3-3x-12y+10)=-8$$

Define $f_y(x)=x^3+y^3-3x-12y+10$. Consider it to be a parametrized function in one variable. You can find its maximum and minimum values (e.g. by solving $f_y'(x)=0$) in dependence of $y$. This will give you a function $x(y)$. To avoid confusion, you might want to rename the variables. Then you minimize $x(y)$ w.r.t. $y$. You will have obtained the values $x$ and $y$ that minimize $f$.

There is no (global) maximum and minimum. This function attains all the values of the form $x^3-3x+10$ (when $y=0$), which in not bounded in $\mathbb R$. By continuity, your function attains every value in $\mathbb R$.

Take the partial derivatives, $f_{x}(x,y)$ and $f_{y}(x,y)$, then find the critical points for each of these such that $f_{x}(x,y) = 0$ or $f_{y}(x,y) =0$ . You'll then need to find second order partial derivatives. Let the point $c$ fulfill the condition: $$c = f_{xx}(a,b)f_{yy}(a,b) - f_{xy^2}(a,b)$$

For c > 0, and $f_{xx}(x,y) > 0$, a relative minimum is at the point $(x,y)$.

For c < 0, and $f_{xx}(x,y) < 0$ there exists a relative maximum at the point $(x,y)$.