# 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?

You reword the issue to make it more practical.

Typically, we formulate these issues as follows: Subject to: (constraint equation) Minimize: (objective function)

Thus, the issue arises:

Now solve a Lagrange multiplier problem as you normally would.

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To use Lagrange multipliers to find the point (a, b) on the graph y = e^(3x) where the value ab is as small as possible, you would set up the following system of equations:

- The objective function: f(a, b) = ab
- The constraint equation: g(a, b) = e^(3a) - b = 0

Next, you would form the Lagrangian function L(a, b, λ) = ab - λ(e^(3a) - b).

Then, you would find the partial derivatives of L with respect to a, b, and λ, and set them equal to zero:

- ∂L/∂a = b(3e^(3a)) - 3λe^(3a) = 0
- ∂L/∂b = a - λ = 0
- ∂L/∂λ = e^(3a) - b = 0

After solving this system of equations, you will find values for a, b, and λ. These values will correspond to the point (a, b) on the graph y = e^(3x) where the value ab is as small as possible.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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