Question-7
Suppose \(x+y=16\). What is the value of \(x y\) when \(x^3+y^3\) is minimum?
\(64\)
We have: \[ \underset{x.y}{\min} \ x^{3} +y^{3} \] subject to: \[ x+y=16 \] This is a constrained optimization problem in two variables. But we can turn it into an unconstrained optimization problem in one variable by using the constraint:
\[ \underset{x}{\min} \ x^{3} +( 16-x)^{3} \] From here, we proceed as usual:
\[ f( x) =x^{3} +( 16-x)^{3} \]
Computing the first and second derivatives:
\[ \begin{aligned} f^{\prime }( x) & =3x^{2} -3( 16-x)^{2}\\ & =3( x+16-x)( x-16+x)\\ & =96( x-8) \end{aligned} \]
There is just one critical point, \(\displaystyle x=8\). Evaluating the second partial derivative at this point:
\[ f^{\prime \prime }( 8) =96 >0 \]
Therefore \(\displaystyle x=8\) corresponds to a local minimum. This is in fact a global minimum as \(f\) turns out to be a quadratic function. We can now recover \(\displaystyle y=8\). The value of \(\displaystyle xy\) is \(\displaystyle 64\) when \(\displaystyle x^{3} +y^{3}\) is minimum.
We have: \[ \underset{x.y}{\min} \ x^{3} +y^{3} \] subject to: \[ x+y=16 \] This is a constrained optimization problem in two variables. But we can turn it into an unconstrained optimization problem in one variable by using the constraint:
\[ \underset{x}{\min} \ x^{3} +( 16-x)^{3} \] We first simplify the objective function:
\[ \begin{aligned} f( x) & =x^{3} +( 16-x)^{3}\\ & =x^3 + 16^{3} -3\times 16^{2} \times x+3\times 16\times x^{2} -x^{3}\\ & =3\times 16x^{2} -3\times 16^{2} x+16^{3} \end{aligned} \]
This is a quadratic equation in \(\displaystyle x\) that has a parabola facing upwards. So it has a global minimum, which occurs at \(\displaystyle x=\frac{3\times 16^{2}}{2\times 3\times 16} =8\). We can now recover \(\displaystyle y=8\). The value of \(\displaystyle xy\) is \(\displaystyle 64\) when \(\displaystyle x^{3} +y^{3}\) is minimum.