Answer to Question #107942 in Calculus for annie

Question

Use Lagrange multipliers to find the point (a,b) on the graph of y=e^8x, where the value ab is as small as possible.

P=?

Answer 1

Consider the objective function $f(x,y)=xy$

Let the constraint function be $g(x,y)=y−e^{8x}$

Use Lagrange Multiplier as follows:

$\nabla f(x,y)=\lambda \nabla g(x,y)$

Next, find the gradients both sides as,

$<y,x>=\lambda <-8e^{8x},1>$

Equate both sides to obtain,

$y=−8λe^{8x}$

$x=λ$

Substitute $x$ for $\lambda$ into relation $y=−8λe^{8x}$ to obtain $y=−8xe^{8x}$

Now, substitute $−8xe^{8x}$ for $y$ into constraint equation $y−e^{8x}=0$ and solve for $x$ as,

$−8xe^{8x}−e^{8x}=0$

$e^{8x}(−8x−1)=0$

$−8x−1=0,e^{8x}\ne0$

$x=-\frac{1}{8}$

Plug $x=-\frac{1}{8}$ into relation $y=−8xe^{8x}$ and solve for $y$ as,

$y=−8(-\frac{1}{8})e^{8(-\frac{1}{8})}=e^{-1}$

Therefore, the required point $(a,b)$ on the graph of $y=e^{8x}$ is $(-\frac{1}{8},\frac{1}{e})$