Final answer:
To minimize the function Q=4x^2+2y^2 given x+y=6, you can substitute y=6-x into Q and then find the derivative to determine the minimum value, which is when x=2 and y=4, resulting in Q=48.
Step-by-step explanation:
The goal is to minimize Q=4x^2+2y^2, subject to the constraint that x+y=6. To solve this optimization problem, we can use a method such as substitution or Lagrange multipliers.
In this case, substitution is a straightforward method. First, we solve the constraint equation for one variable, for example, y=6-x.
Then we substitute this expression for y in the objective function Q, resulting in a single-variable function Q(x).
After substitution, we get Q(x)=4x^2 + 2(6-x)^2.
This can be expanded and simplified to 4x^2 + 2(36 - 12x + x^2)
= 4x^2 + 72 - 24x + 2x^2.
Combining like terms gives us Q(x)=6x^2 - 24x + 72.
Next, we take the derivative of Q with respect to x to find the critical points:
dQ/dx = 12x-24.
Setting this equal to zero gives us x=2.
Since we know y=6-x, we find y=4.
Substituting these values back into the original function yields the minimized value of Q as :
Q=4(2)^2 + 2(4)^2
= 16+32
=48.