# If M and m be the maximum and minimum value of f (x,y) = 7x^2+4xy +3y^2 subjected to the condition x^2+y^2 =1 Find [(M+m)/2]?

Mar 15, 2017

$\frac{m + M}{2} = 5$

#### Explanation:

Changing to polar coordinates using the pass transformations

$\left\{\begin{matrix}x = r \cos \theta \\ y = r \sin \theta\end{matrix}\right.$

we have

$f \left(x , y\right) \to f \left(r , \theta\right) = {r}^{2} \left(7 {\cos}^{2} \theta + 4 \sin \theta \cos \theta + 3 {\sin}^{2} \theta\right)$

and the condition

${x}^{2} + {y}^{2} = 1 \to {r}^{2} = 1$ then the transformed problem is

$\left\{\min , \max\right\} f \left(1 , \theta\right) = \left\{\min , \max\right\} \left(7 {\cos}^{2} \theta + 4 \sin \theta \cos \theta + 3 {\sin}^{2} \theta\right)$

but after simplifications

$7 {\cos}^{2} \theta + 4 \sin \theta \cos \theta + 3 {\sin}^{2} \theta = 5 + 2 \left(\sin \left(2 \theta\right) + \cos \left(2 \theta\right)\right)$

so

$f \left(1 , \theta\right) = 5 + 2 \left(\sin \left(2 \theta\right) + \cos \left(2 \theta\right)\right)$

The stationary points condition is

$\frac{\mathrm{df}}{d \theta} = 4 \left(\sin \left(2 \theta\right) + \cos \left(2 \theta\right)\right) = 0$ and this is attained at

$\theta = \frac{1}{2} \left(\pm \frac{\pi}{4} + 2 k \pi\right)$ for $k = 0 , 1 , 2 , \cdots$

The stationary points qualification is done evaluating

$\frac{{d}^{2} f}{d {\theta}^{2}} = - 8 \left(\sin \left(2 \theta\right) + \cos \left(2 \theta\right)\right)$ resulting in

$\left\{\min , \max\right\} = \left\{8 \sqrt{2} , - 8 \sqrt{2}\right\}$

The values at those points $\theta = \left\{- \frac{\pi}{8} , \frac{\pi}{8}\right\}$ are

$m = 5 - 2 \sqrt{2}$ and $M = 5 + 2 \sqrt{2}$ so

$\frac{m + M}{2} = 5$