Minimizing a Quadratic subject to a Quadratic Constraint

Let the value of f(x,y)=c for point (a,b) such that this point also lies on the second curve. Thus we have 2 equations a²+2ab3b²=c and 4a²+2ab+9b²=100. Since we need to find the max/min value of c, we can imagine in either case both curves must touch each other at the point (a,b)[because the point must be unique and no other point must satisfy the condition within the close neighborhood of the point, otherwise it would be a case of cutting and max/min won't be achieved]. This means that the slopes of both curves must be equal at (a,b). That gives us another condition (a+b)/(a+3b)=(4a+b)/(a+9b) [via elementary calculus, just differentiate the curves and equate dy/dx of both curves], which on simplifying turns out to be a²+ab-2b²=0 =>(a-b)(a+2b)=0 => a=b or a=-2b. Now out of the 2 cases, one is for maximum and other for the minimum value.
Case 1, a=b, we have 15b²=100, 6b²=c => c=40 [by eliminating b²].
Case 2, a=-2b, we have 3b²=c, 21b²=100 => c=100/7.
Thus the min value is 100/7.

We first find whether there's any critical points, $\overrightarrow{c}$ inside the domain $D=\{(x,y)| 4x^2+2xy+9y^2=100\}$ , this is equivalent to finding $\nabla f(\overrightarrow{c})=\overrightarrow{0}$ (since $\nabla f(\overrightarrow{c})$ must exist as $f$ is a multivariate polynomial), so $\nabla f=\langle 2x+2y, 6y+2x\rangle=\langle 0,0 \rangle$ which means $\overrightarrow{c}=\langle x,y \rangle=\langle 0,0 \rangle$ but it does satisfy the given constraint.

Now, we compute the extrema of $f$ on $\partial D$ , the boundary of $D$ , using the method of Lagrange Multipliers, let $g(x,y)=4x^2+2xy+9y^2-100=0$ as our constraint, we have $\nabla f=\lambda \nabla g$ $\langle 2x+2y, 6y+2x\rangle = \lambda \langle 8x+2y, 2x+18y\rangle.$ It is obvious that $\nabla g \ne \overrightarrow{0}$ , otherwise we have $(x,y)=(0,0)$ which doesn't satisfy our constraint. Also, if $\lambda = 0$ , then we also get $(x,y)=(0,0)$ which also doesn't satisfy our constraint.

Then, eliminating $\lambda$ we have $(2x+2y)(18y+2x)=(8x+2y)(6y+2x) \implies x^2+xy=2y^2 \implies (2x+y)^2=9y^2 \implies x=y \ \ \textrm{or} \ \ x=-2y.$ If $x=y$ , then from $g$ , we have $15x^2=100$ , $x^2=\frac{20}{3}$ , so $f=6x^2=40$ . If $x=-2y$ , then from $g$ , we have $21y^2=100$ , $y^2=\frac{100}{21}$ , so $f=3y^2=\frac{100}{7}$ .

Thus, $\boxed{f_{max}=40 \ \textrm{and} \ f_{min}=\frac{100}{7}}$ , $p+q=100+7=\boxed{107}$ .

If we define the symmetric matrices $A \; = \; \left(\begin{array}{cc} 1 & 1 \\ 1 & 3\end{array}\right) \hspace{2cm} B \; =\; \left(\begin{array}{cc} 4 & 1 \\ 1 & 9 \end{array}\right)$ then we want to minimize $f(\mathbf{x}) = \mathbf{x}^TA\mathbf{x}$ subject to the condition $g(\mathbf{x}) = \mathbf{x}^TB\mathbf{x}=100$ . Find an orthogonal matrix $P$ and real numbers $u,v$ such that $B \; = \; P^T\left(\begin{array}{cc} u & 0 \\ 0 & v \end{array}\right)P$ Since $g(x,y) \; = \; 3x^2 + (x+y)^2 + 8y^2$ , it is clear that $g(x,y)$ is positive definite and hence $u,v > 0$ . Thus, if we define $Q \; = \; \left(\begin{array}{cc} \sqrt{u} & 0 \\ 0 & \sqrt{v}\end{array}\right) P$ then $Q$ is a nonsingular matrix and $B = Q^TQ$ . If we change coordinates, and define $\mathbf{X} = Q\mathbf{x}$ , then we are trying to minimize $\mathbf{X}^T(Q^{-1})^TAQ^{-1}\mathbf{X} \hspace{0.5cm} \mbox{subject to} \hspace{0.5cm} \mathbf{X}^T\mathbf{X} = 100$ Since $(Q^{-1})^TAQ^{-1}$ is symmetric, we can find an orthogonal matrix $S$ and reals $a < b$ such that $S^T(Q^{-1})^TAQ^{-1}S \; = \; \left(\begin{array}{cc} a & 0 \\ 0 & b \end{array}\right)$ Changing coodinates once again, defining $\binom{\xi_1}{\xi_1} = S^T\mathbf{X}$ , we end up trying to minimize $a\xi_1^2 + b\xi_2^2$ subject to the restraint $\xi_1^2 + \xi_2^2 = 100$ , and so the minimum value is $100a$ , and the maximum value is $100b$ . Now $\begin{aligned} (t-a)(t-b) & = \; \mathrm{det}\big(tI_2 - S^T(Q^{-1})TAQ^{-1}S\big) \; = \; \mathrm{det}\big(S^T(tI_2 - (Q^{-1})^TAQ^{-1})S\big) \; = \; \mathrm{det}\big(tI_2 - (Q^{-1})^TAQ^{-1}\big) \\ & = \; \mathrm{det}\big((Q^{-1})^T(tQ^TQ - A)Q^{-1}\big) \; = \; \big(\mathrm{det}Q)^{-2}\mathrm{det}(tB-A) \end{aligned}$ and hence we need to solve $0 \; = \; \mathrm{det}(tB-A) \; =\; 35t^2 - 19t + 2 \; =\; (5t-2)(7t-1)$ and hence $a = \tfrac17$ , while $b = \tfrac25$ . Thus the minimum value is $\tfrac{100}{7}$ , making the answer $\boxed{107}$ , and the maximum value is $40$ .