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Lagrange Multipliers: Maximum and Minimum of f(x, y) = 3x + 4y on a Circle
Problem. Find the maximum and minimum values of \(f(x,y)=3x+4y\) on the circle \(x^2+y^2=1\).
Video explanation
Geometric picture
Step 1: Auxiliary function
Constraint: \(g(x,y)=x^2+y^2-1=0\). Form
$$F(x,y,\lambda)=3x+4y+\lambda\,(x^2+y^2-1)$$
Step 2: Partial derivatives
$$F_x=3+2\lambda x,\qquad F_y=4+2\lambda y$$
Step 3: Stationary points
Set \(F_x=0\) and \(F_y=0\):
$$\frac{3}{x}=-2\lambda \quad(3),\qquad \frac{4}{y}=-2\lambda \quad(4)$$
Equating (3) and (4):
$$\frac{3}{x}=\frac{4}{y}\;\Rightarrow\; y=\frac{4x}{3}$$
Substitute into the constraint \(x^2+y^2=1\):
$$x^2+\frac{16x^2}{9}=1\;\Rightarrow\;\frac{25x^2}{9}=1\;\Rightarrow\;x=\pm\frac{3}{5}$$
$$x=\tfrac35\Rightarrow y=\tfrac45,\qquad x=-\tfrac35\Rightarrow y=-\tfrac45$$
Stationary points: \(\left(\tfrac35,\tfrac45\right)\) and \(\left(-\tfrac35,-\tfrac45\right)\).
Step 4: Evaluate f
$$f\!\left(\tfrac35,\tfrac45\right)=\tfrac95+\tfrac{16}5=5,\qquad f\!\left(-\tfrac35,-\tfrac45\right)=-\tfrac95-\tfrac{16}5=-5$$
Maximum value5at (3/5, 4/5)
Minimum value−5at (−3/5, −4/5)
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