# Help with Lagrange Multipliers

• Apr 4th 2013, 12:21 PM
Confucius
Help with Lagrange Multipliers
I am new to the forums, so please forgive any errors in formatting a request for help. Thanks to everyone who views this!
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Here is the question:
"Use Lagrange multipliers to find the point (a, b) on the graph y=ex , where the value ab is as small as possible."

I am unsure how to solve this. Any help would be greatly appreciated since I would like to actually understand the solution and how it is solved!

Once again, thank you!
• Apr 4th 2013, 04:12 PM
Soroban
Re: Help with Lagrange Multipliers
Hello, Confucius!

Quote:

Use Lagrange multipliers to find the point (x, y) on the graph y=ex , where the value xy is as small as possible.

We want to minimize: $f(x,y) \,=\,xy$ subject to the constraint: $e^x - y \,=\,0$

We have: . $F(x,y,\lambda) \;=\;xy + \lambda(e^x-y)$

Set the partial derivatives equal to zero, and solve.

. . $\begin{array}{cccccccc}F_x &=& y + \lambda e^x &=& 0 & [1] \\ F_y &=& x - \lambda &=& 0 & [2] \\ F_{\lambda} &=& e^x-y &=& 0 &[3] \end{array}$

From [2]: . $\lambda \,=\,x$

Substitute into [1]: . $y + xe^x \:=\:0 \quad\Rightarrow\quad y \:=\:-xe^x$

Substitute into [3]: . $e^x - (-xe^x) \:=\:0 \quad\Rightarrow\quad e^x + xe^x \:=\:0$

. . . . . . . . . . . . . . . . . $e^x(1+x) \:=\:0 \quad\Rightarrow\quad x \:=\:-1$

Substitute into [3]: . $e^{-1} - y \:=\:0 \quad\Rightarrow\quad y \:=\:\tfrac{1}{e}$

Therefore: . $(x,y) \;=\;\left(-1,\,\frac{1}{e}\right)$
• Apr 5th 2013, 07:27 AM
Confucius
Re: Help with Lagrange Multipliers
Thank you very much!