Question

Discuss the maxima and minima of the function f(x,y)=x⁴+2x²y-x²+3y²

Answer 1

Answer:

Thus, f(x, y) attains minimum $\frac{-3}{8}$ at $\left(\pm \frac{\sqrt{3}}{2}, \frac{-1}{4}\right)$.

Step-by-step explanation:

Step 1: Solution Let $f(x, y)=x^4+2 x^2 y-x^2+3 y^2$

We have $f x=4 x^3+4 x y-2 x$

and f y=2 x^2+6 y

Then f x=0 and f y=0 implies

$2 x\left(2 x^2+2 y-1\right)=0$ and$2\left(x^2+3 y\right)=0$

i.e., x=0 or 2 x^2+2 y-1=0 and x^2+3 y=0

which is same as

$\left\{x=0\right.$and $\left.x^2+3 y=0\right\}$or$\left\{2 x^2+2 y-1=0\right.$ and $\left.x^2+3 y=0\right\}$

i.e., x=0 and y=0.

where x^2=-3 y

$\begin{aligned}& \therefore \quad 2 x^2+2 y-1=0 \\& 2(-3 y)+2 y-1=0\end{aligned}$

which implies

$y=\frac{-1}{4}$

Hence,$x^2=\frac{3}{4}$ or$x=\pm \frac{\sqrt{3}}{2}$

Step 2: Hence, the critical values are $(0,0)$,

$\left(\frac{\sqrt{3}}{2}, \frac{-1}{4}\right) \text { and }\left(\frac{-\sqrt{3}}{2}, \frac{-1}{4}\right)$

Further

$\begin{aligned}& A=f_{x x}=12 x^2+4 y-2, \\& B=f_{x y}=4 x, C=f_{y y}=6\end{aligned}$

(i) At (0,0), A=-2, B=0,

C=6 \text { and } A C-B^2=-12<0

Hence, there is neither a maximum nor

a minimum at (0,0)

(ii) At$\left(\frac{\sqrt{3}}{2}, \frac{-1}{4}\right)$,

$\begin{aligned}& A=12 \cdot \frac{3}{4}+4\left(\frac{-1}{4}\right)-2=6 \\& B=4 \cdot \frac{\sqrt{3}}{2}=2 \sqrt{3}, C=6\end{aligned}$

Then,$\quad A C-B^2=6(6)-(2 \sqrt{3})^2$ $=24 > 0$ and $A=6 > 0$

$\therefore f(x, y)$ has a minimum at $\left(\frac{\sqrt{3}}{2}, \frac{-1}{4}\right)$

Step 3: Hence, f(x, y) attains its minimum

value at $\left(\frac{\sqrt{3}}{2}, \frac{-1}{4}\right)$

Also, minimum $f(x, y)=\left(\frac{-3}{8}\right)$

(iii) Similarly, at $\left(\frac{-\sqrt{3}}{2}, \frac{-1}{4}\right)$ f(x, y) attains its minimum.

Thus, f(x, y) attains minimum $\frac{-3}{8}$ at $\left(\pm \frac{\sqrt{3}}{2}, \frac{-1}{4}\right)$.

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