MHB Solve Lagrange Multipliers Problem w/ e^(9x) - Find (a,b)

[ajkess1994]

I have been stumped on this problem and I am probably overthinking it as usual. The problem deals with the Lagrange Multipliers:

Use Lagrange multipliers to find the point (a,b) on the graph of y=e^(9x), where the value (a,b) is as small as possible.

I have found the gradient for both when it's set up as [e^(9x) - y = 0], but now I don't know what to do from here. Would someone please be available to help me?

 

[MarkFL]

I have been stumped on this problem and I am probably overthinking it as usual. The problem deals with the Lagrange Multipliers:

Use Lagrange multipliers to find the point (a,b) on the graph of y=e^(9x), where the value (a,b) is as small as possible.

I have found the gradient for both when it's set up as [e^(9x) - y = 0], but now I don't know what to do from here. Would someone please be available to help me?

The objective function is "the value (a,b)" which I interpret to mean the distance from the origin to a point on the curve, and we can use the square of this distance for our objective:

$$f(x,y)=x^2+y^2$$

Subject to the constraint:

$$g(x,y)=e^{9x}-y=0$$

Using Lagrange, we obtain the system:

$$2x=\lambda\left(9e^{9x}\right)$$

$$2y=\lambda(-1)$$

This system implies:

$$\frac{2x}{9e^{9x}}=-2y$$

Or:

$$y=-\frac{x}{9e^{9x}}$$

Now, plugging this into the constraint yields:

$$e^{9x}+\frac{x}{9e^{9x}}=0$$

Or:

$$9e^{18x}+x=0$$

Using a numeric root finding method, we find:

$$x\approx-0.20902833541170100144$$

And so:

$$y\approx0.15239872243985548342$$

We then compute:

$$f(-0.20902833541170100144,0.15239872243985548342)=0.06691821560628668613644254416134077697$$

To ensure we have a minimum, we can use another point on the constraint, such as \((0,1)\), and we find:

$$f(0,1)=1$$

Thus, we may state:

$$f_{\min}\approx0.06691821560628668613644254416134077697$$

Here is a diagram of the solution:

View attachment 8574

 

[ajkess1994]

Thank you for helping me with this problem, enjoy the rest of Veteran's Day