Question

use lagrange multipliers to find the indicated extrema, assuming that x and y are positive.
minimize ( f(x,y)=x^{2}+y^{2} )
constraint: ( x + 2y - 20 = 0 )
( f(quad)=quad )

Explanation:

Step1: Set up the Lagrangian function

Let \(g(x,y)=x + 2y-20\). The Lagrangian function \(L(x,y,\lambda)=x^{2}+y^{2}+\lambda(x + 2y-20)\)

Step2: Take partial derivatives

\(\frac{\partial L}{\partial x}=2x+\lambda = 0\), so \(\lambda=-2x\)
\(\frac{\partial L}{\partial y}=2y + 2\lambda=0\), substitute \(\lambda=-2x\) into it, we get \(2y+2(-2x)=0\), which simplifies to \(y = 2x\)
\(\frac{\partial L}{\partial\lambda}=x + 2y-20=0\)

Step3: Solve the system of equations

Substitute \(y = 2x\) into \(x + 2y-20=0\). Then \(x+2(2x)-20=0\), \(x + 4x-20=0\), \(5x=20\), \(x = 4\)
Since \(y = 2x\), then \(y=8\)

Step4: Calculate the minimum value of \(f(x,y)\)

\(f(x,y)=x^{2}+y^{2}\), substitute \(x = 4\) and \(y = 8\) into it. \(f(4,8)=4^{2}+8^{2}=16 + 64=80\)

Answer:

\(f(4,8)=80\)