Optimization in two variables and Lagrange multipliers, question 4

Topic: Optimization in two variables and Lagrange multipliers
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §16.2Key results

A company sells $x$ thousand wired and $y$ thousand wireless headphones a month. Its monthly profit, in thousands of dollars, is

P(x, y) = 36x + 24y - 2x^{2} - xy - 2y^{2} - 50.

Find the largest monthly profit, in thousands of dollars.

Your answer

A whole number.

How sure are you?

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Theory UBC Math 100 §16.2Key results

Worked solution

Strategy

Find the critical point of P, confirm with the second derivative test that it is a maximum, and evaluate P there.

Working

$P_x = 36 - 4x - y = 0, \qquad P_y = 24 - x - 4y = 0$
The −xy term contributes −y to P_x and −x to P_y.
$y = 36 - 4x \quad\Longrightarrow\quad 24 - x - 4(36 - 4x) = 0 \quad\Longrightarrow\quad 15x = 120$
Solve the first equation for y and substitute into the second.
$x = 8, \qquad y = 36 - 32 = 4$
The only critical point.
$P_{xx} = -4, \quad P_{yy} = -4, \quad P_{xy} = -1, \qquad D = 16 - 1 = 15 > 0$
D > 0 with P_xx < 0: a local maximum. P is a quadratic that curves down in every direction, so this is also its largest value overall.
$P(8, 4) = 288 + 96 - 128 - 32 - 32 - 50 = 142$
Evaluate P at the maximum, every term included.

Answer

P_{\max} = 142 \ \text{(thousand dollars) at } (8, 4)

Check

A thousand more or fewer of either kind, at (9, 4), (7, 4), (8, 5) or (8, 3), brings the profit down to 140, below 142.

Where this goes wrong

Dropping the −xy term when differentiating, so P_x = 36 − 4x and P_y = 24 − 4y. That moves the critical point to (9, 6), where the profit is only 130. The term says the two products compete for the same buyers, and it belongs in both partial derivatives.