How do you use lagrange multipliers to find the point (a,b) on the graph #y=e^(3x)# where the value ab is as small as possible?

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Answer 1 · 2015-03-06T13:15:30

You rewrite the problem in more useful form.

We usually express these problems as:
Minimize: (objective function)
Subject to: (constraint equation)

You want to minimize the product of two number, so the objective function is #f(x,y)=xy#.
The constraint is #y=e^(3x)#.

So the problem becomes:

Minimize: #f(x,y)=xy#
Subject to: #e^(3x)-y=0#

(If you prefer, you could use constraint #y-e^(3x)=0#)

Now, proceed as usual for a Lagrange multiplier problem.