Kunal's question at Yahoo Answers regarding Lagrange multipliers

[MarkFL]

Here is the question:

Use Lagrange multipliers to find the point (a,b) on the graph of y=e^{4 x}, where the value ab is minimal?

Use Lagrange multipliers to find the point (a,b) on the graph of y=e^{4 x}, where the value ab is as small as possible.

I know how to use lagrange multipliers but it isn't working when I follow the formula.

Please help,

Thanks.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Kunal,

We are given the objective function:

\(\displaystyle f(a,b)=ab\)

Subject to the contraint:

\(\displaystyle g(a,b)=b-e^{4a}=0\)

Using Lagrange multipliers, we obtain the system:

\(\displaystyle b=\lambda\left(-4e^{4a} \right)\)

\(\displaystyle a=\lambda(1)\)

And so we find:

\(\displaystyle \lambda=a=-\frac{b}{4e^{4a}}\implies b=-4ae^{4a}\)

Now, substituting for $b$ into the constraint, there results:

\(\displaystyle -4ae^{4a}-e^{4a}=0\)

Divide through by $-e^{4a}\ne0$:

\(\displaystyle 4a+1=0\implies a=-\frac{1}{4}\)

Hence:

\(\displaystyle b=e^{-1}=\frac{1}{e}\)

And so we find the value of the objective function at this point is:

\(\displaystyle f\left(-\frac{1}{4},\frac{1}{e} \right)=-\frac{1}{4e}\)

If we observe that at another point on the constraint, such as $\left(0,1 \right)$, we have:

\(\displaystyle f(0,1)=0\)

Since this is greater than the objective function's value at the critical point we found, we can then conclude with assurance that:

\(\displaystyle f_{\min}=-\frac{1}{4e}\)