Find the minimum of $Q=4 x^{2}+3 y^{2}$ if $x+y=7$.

Step 1 ：We are given a quadratic function in two variables, \(Q=4x^2+3y^2\), and a constraint, \(x+y=7\). We want to find the minimum of \(Q\) subject to this constraint.

Step 2 ：We can use the method of Lagrange multipliers to solve this problem. The Lagrangian is \(L=4x^2+3y^2-\lambda(x+y-7)\), where \(\lambda\) is the Lagrange multiplier.

Step 3 ：We need to solve the system of equations \(\frac{\partial L}{\partial x}=0\), \(\frac{\partial L}{\partial y}=0\), and \(\frac{\partial L}{\partial \lambda}=0\).

Step 4 ：Solving these equations, we get \(\lambda = 24\), \(x = 3\), and \(y = 4\).

Step 5 ：Substituting these values into the equation for \(Q\), we find that \(Q = 84\).

Step 6 ：So, the minimum value of \(Q=4 x^{2}+3 y^{2}\) subject to the constraint \(x+y=7\) is \(\boxed{84}\).