📏📐 Lagrange Multiplier Method | Maximum of f(x,y)=xy on Circle x²+y²=4 ✅ | Example 3 | Applied calculus

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Lagrange Multiplier Example: Maximum of f(x, y) = xy on the Circle x² + y² = 4

CalculusOptimization Example 3 · 4 min read

Problem. Apply the method of Lagrange multipliers to find the maximum of \(f(x,y)=xy\) on the circle \(x^2+y^2=4\).

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VideoGraphStep 1Step 2Step 3Step 4Answer

Video explanation

Geometric picture

Green points (maximum, xy = 2) and red points (minimum, xy = −2) on the circle of radius 2.

Step 1: Auxiliary function

Constraint: \(g(x,y)=x^2+y^2-4=0\).

$$F(x,y,\lambda)=xy+\lambda\,(x^2+y^2-4)$$

Step 2: Partial derivatives

$$F_x=y+2\lambda x,\qquad F_y=x+2\lambda y$$

Step 3: Stationary points

Set \(F_x=0\) and \(F_y=0\):

$$\frac{y}{x}=-2\lambda\quad(3),\qquad \frac{x}{y}=-2\lambda\quad(4)$$

Equating (3) and (4):

$$\frac{y}{x}=\frac{x}{y}\;\Rightarrow\;y^2=x^2$$

Substitute into the constraint \(x^2+y^2=4\):

$$2x^2=4\;\Rightarrow\;x^2=2\;\Rightarrow\;x=\pm\sqrt2,\quad y=\pm\sqrt2$$

Stationary points: \((\sqrt2,\sqrt2),\ (\sqrt2,-\sqrt2),\ (-\sqrt2,\sqrt2),\ (-\sqrt2,-\sqrt2)\).

Step 4: Evaluate f

Point	f(x, y) = xy
(√2, √2)	2
(−√2, −√2)	2
(√2, −√2)	−2
(−√2, √2)	−2

Final answer

Maximum value2at \((\sqrt2,\sqrt2)\) and \((-\sqrt2,-\sqrt2)\)

Quick check: since \(x^2+y^2\ge 2|xy|\), we get \(|xy|\le 2\) on this circle.

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