It is given two positive real numbers \(x,y\) which satisfy \(4x+3y=11\). Find the maximum value of \(f(x,y)=(x+6)(y+7)(3x+2y)\).

Find the maximum value of the following function:

\(f(x,y)=(x+6)(y+7)(3x+2y)\)

To do so, we use the AM-GM inequality:

\(a+b+c\geq3\sqrt[3]{abc}\)

We have:

\(\frac{(x+6)+(y+7)+(3x+2y)}{3}\geq\sqrt[3]{(x+6)(y+7)(3x+2y)}\)

Therefore:

\(\frac{4x+3y+13}{3}\geq\sqrt[3]{f(x,y)}\)

Hence:

\(f(x,y)\leq\left(\frac{4x+3y+13}{3}\right)^3\)

But we know that:

\(4x+3y=11\)

Therefore:

\(f(x,y)\leq\left(\frac{11+13}{3}\right)^3=512\)

Hence, the maximum value of \(f(x,y)=(x+6)(y+7)(3x+2y)\) is:

\(\boxed{512}\)

Solution by: Thin Sokkean