Question

Using Lagranges multiplier method , find the extreme point of
z=xy , subject to x+y=1 .

Answer 1

Answer on Question # 41406 – Math – Integral Calculus

Using Lagrange's multiplier method, find the extreme point of $(x,y) = xy$, subject to $x + y = 1$.

Solution.

Introduce the Lagrange's function:

$$L(x, y, \lambda) = xy + \lambda (x + y - 1);$$

If $A(x_0, y_0)$ is an extreme point, then $A$ satisfies the following system of equations:

$$\left\{ \begin{array}{l} \frac{\partial L}{\partial x} = 0 \\ \frac{\partial L}{\partial y} = 0 \\ \frac{\partial L}{\partial \lambda} = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y + \lambda = 0 \\ x + \lambda = 0 \\ x + y - 1 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = -\lambda \\ x = -\lambda \\ -2\lambda - 1 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = \frac{1}{2} \\ x = \frac{1}{2}; \\ \lambda = -\frac{1}{2} \end{array} \right.$$

Hence, $A\left(\frac{1}{2}, \frac{1}{2}\right)$ is an extreme point.

Answer.

$$A \left(\frac{1}{2}, \frac{1}{2}\right).$$

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