3. Le 25 décembre au matin, Elianne et sa soeur jumelle Maxim down nesoudre une en laissée par le père Noël pour avoir le droit de deboller leur premier codcou Deux-tu le
donner un coup de main?

x = 6, y =3, and the minimum = 45

Step-by-step explanation:

There are plenty of ways to solve this problem.

The Lagranian multiplier method is chosen (it is simple to apply in this case).

Given: A = x^2 + y^2

Find: x and y that satisfy 4x + 2y = 30 or 4x + 2y - 30 = 0 and A is minimum.

This is the same as finding x, y, and lambda that make the following L minimum.

L= x^2 + y^2 + lambda(4x + 2y - 30)

To find x, y, and lambda, we take the derivative of L with respect to them, set the derivative to 0, and solve the equations.

L'(x) = 2x + 4(lambda) = 0    (1)

L'(y) = 2y + 2(lambda) = 0    (2)

L'(lambda) = 4x + 2y - 30 = 0   (3)

Multiply (2) by 2 and subtract to (1), we have:

4y - 2x = 0  (*)

Combine (*) with (3) we have:

4y - 2x = 0

4x + 2y - 30 = 0

Multiply the 1st equation by 2 and add to the 2nd equation:

=>  10y - 30 = 0

=> y = 3

=> x = 4y/2 = 4(3)/2 = 6 (as (*))

=> lambda = -y = -3  (as (2))

Substitute x = 6, y = 3, and lambda = -3 into L, or substitute x = 6 and y = 3 into A to get the minimum.

Both ways would lead to the same answer.

Check:

L = x^2 + y^2 + lambda(4x + 2y - 30)

  = 6^2 + 3^2 + (-3)[4(6) + 2(3) - 30]

  = 36 + 9 + (-3)(0)

  = 45

A = x^2 + y^2

   = 36 + 9

   = 45

Hope this helps!