Optimization in two variables and Lagrange multipliers, question 9

Topic: Optimization in two variables and Lagrange multipliers
Difficulty: Challenging
Suggested time: 6 min
Marks: 3
Theory: UBC Math 100 §16.3Key results

A bakery’s daily output is $Q = 2\sqrt{xy}$ trays, where $x$ is hours of baker time at $25 an hour and $y$ is hours of oven time at $16 an hour. It must make 40 trays a day. Use Lagrange multipliers to find the least daily cost of doing so.

Your answer

A whole number of dollars.

How sure are you?

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Theory UBC Math 100 §16.3Key results

Worked solution

Strategy

Cost is the objective and the output target is the constraint. The Lagrange conditions fix the ratio of the two inputs, and the target then fixes their sizes.

Working

$25 = \lambda\sqrt{\frac{y}{x}}, \qquad 16 = \lambda\sqrt{\frac{x}{y}}, \qquad 2\sqrt{xy} = 40$
C_x = λg_x and C_y = λg_y with g = 2√(xy), plus the target.
$\frac{25}{16} = \frac{\sqrt{y/x}}{\sqrt{x/y}} = \frac{y}{x} \quad\Longrightarrow\quad y = \frac{25}{16}x$
Divide the first equation by the second; λ cancels.
$xy = 400 \ \Rightarrow\ \frac{25}{16}x^{2} = 400 \ \Rightarrow\ x^{2} = 256 \ \Rightarrow\ x = 16,\ y = 25$
The target 2√(xy) = 40 means xy = 400. Keep the positive root.
$C = 25(16) + 16(25) = 400 + 400 = 800$
Along xy = 400 the cost grows without bound as either input shrinks toward 0, so this single candidate is the minimum.

Answer

C_{\min} = \$800 \ \text{with}\ x = 16,\ y = 25

Check

Other ways to make 40 trays cost more: 20 hours of each costs $820, and 10 baker hours with 40 oven hours cost $890.

Where this goes wrong

Dividing the conditions the wrong way round, so that y/x = 16/25. That gives x = 25 and y = 16, which still makes 40 trays but costs $881. A quick check: the cheaper input, oven time, should be used more.