Question

To find the maximum and the minimum points of the given function.

$\mathbf{4}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{xy}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}$

Short answer

The maximum point of the given function is (0,1) .

Step-by-step solution

Given data

A function is given as $4+\mathrm{x}+\mathrm{y}-{\mathrm{x}}^2-\mathrm{xy}-\frac{1}{2}{\mathrm{y}}^2$.

Concept of maximum and minimum points of a function

To evaluate the maximum and minimum points of a functionthe following points should be noted:

1. ${\mathbf{f}}_{\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{=}{\mathbf{f}}_{\mathbf{y}}$At(a,b)

2. (a,b) is a minimum point if localid="1658833120511" ${\mathit{f}}_{\mathbf{x}\mathbf{x}}\mathbf{>}\mathbf{0}\mathbf{,}{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{>}\mathbf{0}$, and localid="1658833144703" ${\mathbf{f}}_{\mathbf{xx}}{\mathbf{f}}_{\mathbf{yy}}\mathbf{>}{\mathbf{f}}_{\mathbf{xy}}^{\mathbf{2}}$

3.(a,b)is a maximum point if localid="1658833202518" ${\mathit{f}}_{\mathbf{xx}}\mathbf{<}\mathbf{0}\mathbf{,}{\mathit{f}}_{\mathbf{yy}}\mathbf{<}\mathbf{0}$, and ${\mathbf{f}}_{\mathbf{xx}}{\mathbf{f}}_{\mathbf{yy}}\mathbf{>}{\mathbf{f}}_{\mathbf{xy}}^{\mathbf{2}}$

4.(a,b) is neither a maximum nor a minimum point if ${\mathbf{f}}_{\mathbf{xx}}{\mathbf{f}}_{\mathbf{yy}}\mathbf{<}{\mathbf{f}}_{\mathbf{xy}}^{\mathbf{2}}$that is ${\mathbf{f}}_{\mathbf{yy}}$and ${\mathbf{f}}_{\mathbf{xx}}$are of opposite sign.

Here,role="math" localid="1658833334348" ${\mathbf{f}}_{\mathbf{x}}\mathbf{,}{\mathbf{f}}_{\mathbf{y}}$are the first-order partial derivatives and ${\mathbf{f}}_{\mathbf{xx}}\mathbf{,}{\mathbf{f}}_{\mathbf{yy}}\mathbf{,}\mathbf{and}\mathbf{}{\mathbf{f}}_{\mathbf{xy}}$are the second-order partial derivatives.

Differentiate the equation of f(x,y) 

Consider the given function as follows:

$f\left(x,y\right)=4+x+y-x^2-xy-\frac{1}{2}{y}^2.......\left(1\right)$

Differentiate (1) partially with respect to x as shown below.

role="math" localid="1658833697925" $$\begin{gathered} f_x\left(x,y\right)=\frac{\partial }{\partial x}\left(4+x+y-x^2-xy-\frac{1}{2}{y}^2\right) \\ {f}_x\left(x,y\right)=\frac{\partial }{\partial x}4+\frac{\partial }{\partial x}x+\frac{\partial }{\partial x}y-\frac{\partial }{\partial x}{x}^2-\frac{\partial }{\partial x}xy-\frac{\partial }{\partial x}\frac{1}{2}{y}^2 \\ {f}_x\left(x,y\right)=1-2x-y......\left(2\right) \end{gathered}$$

Differentiate (1) partially with respect to as shown below:

$$\begin{gathered} f_x\left(x,y\right)=\frac{\partial }{\partial y}\left(4+x+y-x^2-xy-\frac{1}{2}{y}^2\right) \\ {f}_x\left(x,y\right)=\frac{\partial }{\partial y}4+\frac{\partial }{\partial y}x+\frac{\partial }{\partial y}y-\frac{\partial }{\partial y}{x}^2-\frac{\partial }{\partial y}xy-\frac{\partial }{\partial y}\frac{1}{2}{y}^2 \\ {f}_x\left(x,y\right)=1-x-y......\left(3\right) \end{gathered}$$

Solve for the value of x and y

Equate equation (2) with zero as follows:

$$\begin{gathered} f_x\left(x,y\right)=1-2x-y \\ 0=1-2x-y...........\left(4\right) \end{gathered}$$

Now, equate equation (3) with zero as follows:

$$\begin{gathered} f_x\left(x,y\right)=1-x-y \\ 0=1-x-y\left(5\right) \end{gathered}$$

Solve the equations (4) and (5), to get $x=0,y=-1$.

The obtained point is (0,1).

Calculation for the minimum points of the function  

Now, differentiate (2) and (3) again as shown below.

$f_{xx}(x,y)=-2$……. (6)

$f_{yy}(x,y)=-1$……. (7)

$f_{xy}(x,y)=-1$……. (8)

Now, use the value obtained in the equation (6),(7),(8) to check the maximum and minimum points.

At (0,1), $f_{xx}<0,\hspace{0.33em}{f}_{yy}<0$and $f_{xx}{f}_{yy}>{f_{xy}}^2$.

Therefore, the point (0,1) is the maximum point of the given function.