Solve the optimization problem.
Minimize F = x^2 + y^2 with x + 2y = 15.
Thank You for the help!!
x = 15 - 2y
x^2 = 225 - 60 y + 4 y^2
F = 5 y^2 -60 y + 225
dF/dy = 10 y - 60
zero when y = 6 then x = 3
is that min or max?
d^2F/dy^2 = 10 so it is min
F here = 9+36 = 45
check ends
if x = 0 y = 7.5 and F = 56 1/4
if y = 0, x=15 and F = 225
ok, done
Wow thank you so much!!!!
To solve the given optimization problem, we'll use a technique called Lagrange multipliers. This method allows us to optimize a function subject to a constraint.
First, we need to set up the problem using Lagrange multipliers. Let's define the Lagrangian function:
L(x, y, λ) = F(x, y) - λ(g(x, y))
where F(x, y) is the objective function we want to minimize (F = x^2 + y^2) and g(x, y) is the constraint (x + 2y = 15). λ is a Lagrange multiplier.
Next, we find the partial derivatives of L with respect to each variable:
∂L/∂x = 2x - λ
∂L/∂y = 2y - 2λ
∂L/∂λ = -(x + 2y - 15)
Setting each partial derivative equal to zero, we can solve for x, y, and λ. First, we solve for x:
2x - λ = 0
=> x = λ/2
Next, we solve for y:
2y - 2λ = 0
=> y = λ
Finally, we solve for λ using the constraint:
x + 2y = 15
=> λ/2 + 2λ = 15
=> 5λ = 30
=> λ = 6
Now that we have the value of λ, we can substitute it back into the equations for x and y:
x = λ/2 = 6/2 = 3
y = λ = 6
Therefore, the solution to the optimization problem is x = 3, y = 6. Substituting these values into the objective function F, we get F = 3^2 + 6^2 = 9 + 36 = 45.
So, the minimum value of F is 45 when x = 3 and y = 6.
I hope this explanation helps you solve optimization problems using Lagrange multipliers in the future. If you have any further questions, feel free to ask!