NBHM 2020 PART A Question 11 Solution

What are the maximum and minimum values of $x + y$ in the region $S = \{(x, y) : x^2 +4y^2 ≤ 1\}$? 
SolutionA convex function on a compact convex set attains its maximum and minimum at an extreme point of the set. So the required maximum and minimum should attain on the boundary $T = \{(x,y): x^2 + 4y^2 = 1\}$. We now use the Lagrange's multiplier to find the maximum and minimum of the given function in the boundary $T$. 

Let $f(x,y) = x + y$ and $g(x,y) = x^2 + 4y^2 -1$. Lagrange's multiplier $\lambda$ is given by $\mathcal L(x,y,\lambda) = f(x,y) + \lambda g(x,y) = (x+y) + \lambda (x^2+4y^2-1)$. To find the required maximum and minimum, we need to find the gradient of $\mathcal L$ and equate it to zero. The gradient of $\mathcal L$ is equal to
$$\nabla(\mathcal{L}) =(\frac{\partial \mathcal {L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\mathcal{L}}{\partial \lambda}) = (1+2\lambda x, 1+8\lambda y, x^2 + 4y^2-1).$$
$\nabla(\mathcal{L}) = 0$ implies that $$1+2\lambda x = 0,$$ $$1+8\lambda y = 0,$$ $$x^2 + 4y^2 -1 = 0.$$ From first two equations we get, $$x = \frac{-1}{2\lambda},$$ $$y = \frac{-1}{8\lambda}$$ Substitute this value in the third equation we get $$\lambda = \pm \frac{\sqrt 5}{4}.$$ Now substitue the values of $\lambda$ in the above two equations we get $$x = \frac{-2}{\sqrt 5},$$ $$y = \frac{-1}{2\sqrt 5}$$
Hence the stationary points are $$(\frac{-2}{\sqrt 5}, \frac{-1}{2\sqrt{5}},\frac{\sqrt{5}}{4})$$ and $$(\frac{2}{\sqrt 5}, \frac{1}{2\sqrt{5}}, \frac{-\sqrt 5}{4}).$$

Substitute the $x$ and $y$ coordinates of stationary points in $f(x,y)$  we get $x+y = \pm \frac{\sqrt 5}{2}$ which is the required maximum and minimum.



Share to your groups:
FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO, SUGGEST PROBLEMS TO SOLVE.

No comments:

Post a Comment

Featured Post

NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...

Popular Posts