Optimization in two variables and Lagrange multipliers, question 8: UBC MATH 100 practice

Question 8 of 14Standard3 marks · 4 minTheory UBC Math 100 §16.3Key results

Use Lagrange multipliers to find the extreme values of $f(x, y) = x^{2} + y^{2}$ on the curve $xy = 4$ . Which statement is correct?

A maximum value of 8, and no minimum value
A minimum value of 8, and no maximum value
A minimum value of 8 and a maximum value of 8
No maximum and no minimum, because the curve is unbounded

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Theory UBC Math 100 §16.3Key results

Worked solution

Strategy

Find the Lagrange candidates, then ask whether each extreme exists at all. The hyperbola xy = 4 is unbounded, so f need not have both a largest and a smallest value on it.

Working

$2x = \lambda y, \qquad 2y = \lambda x, \qquad xy = 4$
The Lagrange conditions with g(x, y) = xy, so g_x = y and g_y = x.
$2x^{2} = \lambda xy = 2y^{2} \quad\Longrightarrow\quad x^{2} = y^{2}$
Multiply the first equation by x and the second by y: both right sides become λxy.
$xy = 4 > 0 \ \Rightarrow\ y = x,\ x^{2} = 4: \qquad (2, 2),\ (-2, -2),\ \ f = 8$
y = −x would make xy negative, so only y = x survives. Both candidates give f = 8.
$f\left(8, \tfrac{1}{2}\right) = 64.25, \qquad f \to \infty \ \text{as}\ x \to 0 \ \text{or}\ x \to \pm\infty$
Along the curve, far out on either branch, f grows without bound, so there is no maximum.
$f_{\min} = 8, \qquad \text{no maximum}$
Because f is large at both ends of each branch, it has a least value on the curve, and that must be at a candidate.

Answer

f_{\min} = 8 \ \text{at}\ (2, 2) \text{ and } (-2, -2); \quad \text{no maximum}

Check

Along the curve, f = x² + 16/x² is 17 at x = 1 and x = 4, more than 8 at x = 2, and grows without bound as x → 0.

Where this goes wrong

Assuming the candidates supply both a maximum and a minimum, so that 8 is both. That is guaranteed only on a closed, bounded constraint such as a circle. On xy = 4 the point (8, 1/2) already gives f = 64.25.