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Determine the maximum and minimum possible values of 5x**2 + 2y**2, under the constraint x**2 + y**2 <= 9. You may assume maxima and minima exist.

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User Lolix
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7.4k points

1 Answer

4 votes

To solve this problem, we will utilize the method of Lagrange multipliers.

First step is to write down the original and constraint functions. In this case:
Function F: 5x^2 + 2y^2
Constraints: x^2 + y^2 - 9 = 0

Next, we differentiate the function F and constraint, first with respect to x, and then to y:

Partial derivatives of F:
with respect to x: F'x = 10x.
with respect to y: F'y = 4y.

Partial derivatives of constraints (g as constraint function):
with respect to x: g'x = 2x.
with respect to y: g'y = 2y.

We can find a scalar (which is denoted as λ) satisfying:
F'x = λ * g'x,
F'y = λ * g'y.

Solving these equations, we get:
10x = λ * 2x => λ = 5
and
4y = λ * 2y => λ = 2

By equating the two values of λ, it turns out that there is no solution, implying that the maxima and minima can only be on the boundary.

The boundary of the constraint x^2 + y^2 = 9 is the circle of radius 3. Substituting x^2 + y^2 = 9 into the original function, we have:

F = 5 * 9 + 2 * (9 - x^2) = 45 + 18 -2 x^2 = 63 - 2x^2.

This is a parabola opening downwards, with maximum at x=0, so the maximum value of F=x^2 + y^2 under the constraint is 63. This maximum value is found at (0,3) and (0,-3).

We can also say that the minimum of F (5x^2 + 2y^2) is 0 as the function is always positive and becomes 0 at the point where x and y equals 0.

So, the minimum and maximum values of the function are 0 and 63 respectively.

answered
User Skippy Fastol
by
8.3k points
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