Discussion
Consider the function \(f: \mathbb{R}^{2} \to \mathbb{R}\) given by
\[ f(x, y) = xy(5x + 3y - 6). \]
Which one of the following statements is NOT true?
Answer: C
The statement that is NOT true is \(f\) has two local minima; it has only one.
Write \(f = 5x^{2}y + 3xy^{2} - 6xy\). Then
\[ \begin{aligned} f_{x} &= y(10x + 3y - 6) \\ f_{y} &= x(5x + 6y - 6) \end{aligned} \]
Setting both to zero gives four critical points: \((0,0)\), \(\left(\frac{6}{5}, 0\right)\), \((0, 2)\), and the solution of \(10x + 3y = 6\), \(5x + 6y = 6\), which is \(\left(\frac{2}{5}, \frac{2}{3}\right)\), with \(a + b = \frac{16}{15}\).
With \(f_{xx} = 10y\), \(f_{yy} = 6x\), \(f_{xy} = 10x + 6y - 6\), the Hessian determinant \(f_{xx}f_{yy} - f_{xy}^{2}\) is \(-36\) at each of the first three points, so they are saddle points. At \(\left(\frac{2}{5}, \frac{2}{3}\right)\) it is \(\frac{20}{3} \cdot \frac{12}{5} - 2^{2} = 12 \gt 0\) with \(f_{xx} \gt 0\): the only local minimum. So statements A, B and D are true.
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