Maxima and Minima Engineering Maths, Btech first year

Maxima and Minima in two variables

You have studied maxima and minima in one variable, here you will get to know how to find maxima and minima in two variables for a function $f(x,y)$.

Step 1 : Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$

Step 2 : Put $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ and solve for $(x,y)\Rightarrow$ Stationary/Critical points

Step 3 : Find $$\frac{{\partial }^{2}f}{\partial x^2}=r=?$$ $$\frac{{\partial }^{2}f}{\partial y^2}=t=?$$ $$\frac{{\partial }^{2}f}{\partial x\partial y}=s=?$$ Step 4 : Find $rt-s^2=?$

Step 5 : Put the points $(x,y)$ in $rt-s^2$

Step 6 : If the value of $rt-s^2$ is :
(a) Positive, then put the points in $r$
$$(a.1) If Positive then the point is Point of Minima
$$(a.2) If Negative then the point is Point of Maxima
(b) Negative, then it is the case of Neither Maxima Nor Minima
(c) equal to 0, then further investigation required

Sample Problem 1

Find maximum and minimum values of function $x^3+y^3-3axy,a>0$

Solution :

$$f(x,y)=x^3+y^3-3axy⟶(1)$$ Partially differentiating w.r.t. $x,y$ $$\frac{\partial f}{\partial x}=3x^2-3ay⟶(2)$$ $$\frac{\partial f}{\partial y}=3y^2-3ax⟶(3)$$ For stationary points, $$\text{Put }\frac{\partial f}{\partial x}=0$$ $$3x^2-3ay=0$$ $$x^2=ay⟶(4)$$ $$\text{Put }\frac{\partial f}{\partial y}=0$$ $$3y^2-3ax=0$$ $$y^2=ax⟶(5)$$ From equation (4) $$x^2=ay\Rightarrow \boxed{{\displaystyle y=\frac{x^2}{a}}}$$ Putting value of $y$ in equation (5) $${\left(\frac{x^2}{a}\right)}^2=ax$$ $$\Rightarrow \frac{x^4}{a^2}=ax$$ If you are thinking to cancel out $x$, don't, because it's a variable, not a constant. $$x^4=a^{3}x$$ $$x(x^3-a^3)=0$$ $$\Rightarrow \boxed{{\displaystyle x=0}}\mathrm{\&}\boxed{{\displaystyle x=a}}$$ Putting values of $x$ in equation (4) $$x^2=ay$$ $$\text{For }x=0,0=ay\Rightarrow \boxed{{\displaystyle y=0}}$$ $$\text{For }x=a,a^2=ay\Rightarrow \boxed{{\displaystyle y=a}}$$ So the stationary points are $(0,0)\mathrm{\&}(a,a)$
Now we will find values of $r,s,t$ $$r=\frac{{\partial }^{2}f}{\partial x^2}=6x\Rightarrow \boxed{{\displaystyle r=6x}}$$ $$s=\frac{{\partial }^{2}f}{\partial x\partial y}=0-3a\Rightarrow \boxed{{\displaystyle s=-3a}}$$ $$t=\frac{{\partial }^{2}f}{\partial y^2}=6y\Rightarrow \boxed{{\displaystyle t=6y}}$$ $$\text{Now }rt-s^2$$ $$=(6x)(6y)-(-3a{)}^2$$ $$=36xy-9a^2$$ $$\text{At point }(0,0)$$ $$rt-s^2=0-9a^2=-9a^2<0[\text{As }a>0]$$ $$\Rightarrow \text{Neither Minima Nor Maxima}$$ $$\text{At point }(a,a)$$ $$rt-s^2=36(a)(a)-9a^2=27a^2>0$$ $$\Rightarrow \text{Point of Minima}$$ Now for the minimum value of the function, put the point of minima in $f(x,y)$ $$f(a,a)=a^3+a^3-3a^3$$ $$=-a^3$$

Sample Problem 2

Find the absolute maximum and minimum values of $$f(x,y)=2+2x+2y-x^2-y^2$$

Solution

Partially differentiating $f(x,y)$ w.r.t. $x,y$ $$\frac{\partial f}{\partial x}=2-2x$$ $$\frac{\partial f}{partialy}=2-2y$$ For stationary points, $$\frac{\partial f}{\partial x}=0$$ $$\Rightarrow 2-2x=0$$ $$\boxed{{\displaystyle x=1}}$$ $$\frac{\partial f}{\partial y}=0$$ $$\Rightarrow 2-2y=0$$ $$\boxed{{\displaystyle y=1}}$$ So the stationary point is (1,1)
Now, $$r=\frac{{\partial }^{2}f}{\partial x^2}=-2$$ $$s=\frac{{\partial }^{2}f}{\partial x\partial y}=0$$ $$t=\frac{{\partial }^{2}f}{\partial y^2}=-2$$ $$rt-s^2=(-2)(-2)-0=4$$ So maxima or minima will be decided by the value of $r$ $$r=-2<0$$ Hence (1,1) is the point of maxima
Maximum value of the function $$=2+2+2-1-1=4$$ $$=\boxed{{\displaystyle 4}}$$