Quadratic Equation

Note that $P(x)=4x^2+6x+4=\left(2x+\frac{3}{2}\right)^2+\frac{7}{4}\geq\frac{7}{4}$ and $Q(y)=(2y-3)^2+16\geq 16$ . This implies $P(x)Q(y)\geq\frac{7}{4} \times 16=28$ . Since it is given that $P(x)Q(y)=28$ , we have $2x+\frac{3}{2}=0 \implies x=-\frac{3}{4}$ and $2y-3=0 \implies y=\frac{3}{2}$ . Hence, $11y-26x=\boxed{36}$ .

We use boundary condition :-

Minimum value of P(x) = -D/4a = 7/4

Minimum value of Q(y) = -D/4a = 16

Minimum value of P(x).Q(y) = 7/4×16 = 28

At P(x) = 7/4 , x= -b/2a = -3/4

At Q(y) = 16 , y= -b/2a = 3/2

11y-26x = 36

These problems are of resonance module