 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability-resit/4/c/solution

Use

F = x^{6} + \frac{3}{2} y^{4}

and

G = x^{4} - 2 y^{2} - 3

x \neq = 0

, so

\nabla G = (4 x^{3}, - 4 y)

6 x^{5} = 4 λ x^{3}, 6 y^{3} = - 4 λ y .

The first gives

λ = \frac{3}{2} x^{2} > 0

. If

y \neq = 0

, the second would give

λ = - \frac{3}{2} y^{2} < 0

, a contradiction. Therefore

y = 0

, and the constraint gives

x^{4} = 3

. The only constrained stationary points are

(\pm 3^{1/4}, 0)

For a global certificate, feasibility implies

x^{4} \geq 3

, and hence

F = (x^{4})^{3/2} + \frac{3}{2} y^{4} \geq 3^{3/2} .

Equality requires

y = 0

and

x^{4} = 3

. Thus

min F = 33, (x, y) = (\pm 3^{1/4}, 0) .

The case split is essential: dividing by

y

would discard precisely the minimizers.

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