Answer to Question #107940 in Calculus for annie

Question

Find the maximum and minimum values of the function f(x,y)=2x^2+3y^2−4x−5 on the domain x2+y2≤225.

What is the maximum value of f(x,y)?


List the point(s) where the function attains its maximum as an ordered pair, such as (-6,3), or a list of ordered pairs if there is more than one point, such as (1,3), (-4,7).

What is the minimum value of f(x,y) ?


List points where the function attains its minimum as an ordered pair, such as (-6,3), or a list of ordered pairs if there is more than one point, such as (1,3), (-4,7).

Answer 1

Given $f(x,y) = 2x^{2}+3y^{2}-4x-5$.

We need to use Lagrange multipliers for the boundary and partial derivatives to determine critical points from the interior. Consider the interior first. Then we have to solve the equations,

$\dfrac{\partial f}{\partial x} = 0\\$ and $\dfrac{\partial f}{\partial y} = 0$. From which we get,

$4x-4=0$ and $6y=0.$ Thus, $(x, y) = (1,0).$

This critical point lies in the region $x^2+y^2<225$ and the value of f at this point is $f(1,0)=2-4-5 = -7.$

On the boundary, we find the extreme values of $f(x,y) = 2x^{2}+3y^{2}-4x-5$ subject to the constraint $g(x,y) = x^{2}+y^{2}-225$ using Lagrange's multipliers.

Let

$F = f + \lambda g \\= 2x^2+3y^2-4x-5+\lambda(x^2+y^2-225)$

To find the critical points we have to solve the equations,

$\dfrac{\partial F}{\partial x} = 0, \dfrac{\partial F}{\partial y} = 0, \dfrac{\partial F}{\partial \lambda} = 0\\$.

$4x-4+2x\lambda = 0\\ 6y+2y\lambda = 0\\ x^2+y^2 = 225$

From the second equation we get, $y=0$ or $\lambda = -3$. If $y = 0$ , then $x = \pm15$ , from the constraint.

When $\lambda = -3$ , we get $x = -2$ and from constraint we get, $y = \pm\sqrt{221}$.

The critical points are $(1,0), (-15,0),(15,0), (-2,-\sqrt{221}), (-2,\sqrt{221})$.

The values at these points are,

$f(1,0) = -7\\ f(-15,0) = 505\\ f(15,0) = 385\\ f(-2,-\sqrt{221}) = 674\\ f(-2,\sqrt{221}) = 674\\$

Therefore, the maximum value is 674 and the minimum value is -7.

The maximum value is attained at the points $(-2, -\sqrt{221}), (-2,\sqrt{221})$ and the minimum value is attained at the point $(1,0).$