Answer:
x = 6, y =3, and the minimum = 45
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!