Steply

Max and min of f(x, y) = 6x + 8y on x^2 + y^2 = 1

Step by step, with the reason for every step · Lagrange multipliers · Calculus 3

Step by step

The question

Question: find the maximum and minimum of f(x, y) = 6x + 8y with x² + y² = 1.

f(x, y) = 6x + 8y, x2 + y2 = 1

1 Set up the equations

At the best point, the level curve of f just touches the curve.

So the two gradients point the same way.

The rule: ∇f = λ∇g.

λ is the Greek letter lambda.

It is just a number.

∇f = λ ∇g

The curve is g = 1, with g(x, y) = x² + y².

g(x, y) = x2 + y2

The x parts: f_x = λ g_x.

f_x is 6, g_x is 2x.

① 6 = λ × 2x

The y parts: f_y = λ g_y.

f_y is 8, g_y is 2y.

② 8 = λ × 2y

And ③: stay on the curve.

③ x2 + y2 = 1

Careful!

The curve's own equation is part of the system.

Without it, we cannot find x and y.

Careful: three equations: never forget ③

2 Solve the equations

From ①: 6 = λ × 2x, so λ is 6/2x.

λ = 62x

From ②: λ is 8/2y.

λ = 82y

Both are λ, so they are equal.

62x = 82y

Cross-multiply and tidy up: y = 4/3x.

y = 43x

Put it into ③.

x² + (4/3x)² = 1

Add the x² terms and divide: x² = 9/25.

x² = 925

Two numbers: x = 3/5 or −3/5.

Then y follows: two points.

x = 3/5 or −3/5

3 Compare the values

At (3/5, 4/5), f is 10.

f(3/5, 4/5) = 10

At (−3/5, −4/5), f is −10.

f(−3/5, −4/5) = −10

The bigger value is the maximum, the smaller one the minimum.

= max 10 at (3/5, 4/5), min −10 at (−3/5, −4/5)

4 Check

Let's check on the xy grid.

The orange circle is the curve ③.

The blue lines are where f is 10 and −10.

Each just touches the circle.

Another point on the circle gives 6: between the two.

Right!

f(1, 0) = 6

Let's look back.

First: set up the equations.

Then: solve the equations.

Then: compare the values.

Last: check.

Answer
= max 10 at (3/5, 4/5), min −10 at (−3/5, −4/5)

Your turn

Try this one yourself, then check:

Find the maximum and minimum of f(x, y) = 3x + 4y, x2 + y2 = 1
Show the answer
max 5 at (3/5, 4/5), min −5 at (−3/5, −4/5)

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