Optimization in two variables and Lagrange multipliers, question 10

Topic: Optimization in two variables and Lagrange multipliers
Difficulty: Capstone
Suggested time: 10 min
Marks: 6
Theory: UBC Math 100 §16.3Key results

A workshop’s output is $Q = x\sqrt{y}$ units, where $x$ is hours of labour at $25 an hour and $y$ is units of material at $50 each. Its budget is $2400. Use Lagrange multipliers to find the maximum output on this budget. Then find the maximum output on a budget of $3750, and give the increase.

Give the increase in maximum output.

Your answer

units

A whole number.

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Theory UBC Math 100 §16.3Key results

Worked solution

Strategy

Lagrange multipliers fix the ratio of the two inputs, and the budget does not enter that ratio. Each budget then fixes the sizes of the inputs, and the increase is the difference of the two maxima.

Working

$\sqrt{y} = 25\lambda, \qquad \frac{x}{2\sqrt{y}} = 50\lambda, \qquad 25x + 50y = B$
Q_x = λg_x, Q_y = λg_y and the budget B, with g(x, y) = 25x + 50y.
$\frac{x/(2\sqrt{y})}{\sqrt{y}} = \frac{50\lambda}{25\lambda} \quad\Longrightarrow\quad \frac{x}{2y} = 2 \quad\Longrightarrow\quad x = 4y$
Divide the second equation by the first; λ cancels, and B never appears.
$B = 2400: \quad 150y = 2400, \quad y = 16, \quad x = 64, \quad Q_{\max} = 64\sqrt{16} = 256$
At either end of the budget line one input is zero and Q = 0, so this candidate is the maximum.
$B = 3750: \quad 150y = 3750, \quad y = 25, \quad x = 100, \quad Q_{\max} = 100\sqrt{25} = 500$
The same ratio on the larger budget.
$500 - 256 = 244$
Both inputs grow by the factor 25/16 and √y by 5/4, so output grows by 125/64: the budget rises by just over half, but output nearly doubles.

Answer

Q_{\max} \text{ rises from } 256 \text{ to } 500, \quad \text{an increase of } 244 \text{ units}

Check

Neighbouring bundles do worse: on $2400, 60 hours and 18 units give 60√18 ≈ 254.6, and on $3750, 110 hours and 20 units give 110√20 ≈ 491.9.

Where this goes wrong

Dividing the conditions the wrong way round, which gives x = y instead of x = 4y and an increase of 122√2 ≈ 173. Substitute back to check: with x = 4y, x/(2√y) = 2√y, which is 50λ when √y = 25λ.