To minimize Q=3x^2+3y^2, we need to solve the system of equations x+y=6 and Q=3x^2+3y^2. Taking the derivative of Q and setting it equal to zero, we find x=3. Substituting x=3 into x+y=6 gives us y=3. Therefore, the minimum value of Q is obtained when x=3 and y=3.
To minimize Q=3x2+3y2, we need to find the values of x and y that satisfy the condition x+y=6. This can be done by solving the system of equations:
x+y=6
Substituting y=6-x into Q=3x2+3y2, we get:
Q=3x2+3(6-x)2
Simplifying and expanding, we have:
Q=3x2+3(36-12x+x2)
Q=3x2+108-36x+3x2
Combining like terms, we get:
Q=6x2-36x+108
To minimize Q, we can take the derivative of Q with respect to x and set it equal to zero:
dQ/dx=12x-36=0
Solving for x, we find x=3.
Substituting x=3 into the equation x+y=6, we get y=3.
Therefore, the minimum value of Q is obtained when x=3 and y=3.