Answer
The point $\left( {a,b} \right)$ on the graph of $y = {{\rm{e}}^x}$ where the value $a b$ is as small as possible occurs at the point $\left( { - 1,{{\rm{e}}^{ - 1}}} \right)$.
Work Step by Step
We have $y = {{\rm{e}}^x}$. Let $f\left( {x,y} \right) = xy$.
Our task is to find two numbers $a$ and $b$ such that $f\left( {a,b} \right) = ab$ is the minimum value subject to the constraint $g\left( {x,y} \right) = {{\rm{e}}^x} - y = 0$.
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {y,x} \right) = \lambda \left( {{{\rm{e}}^x}, - 1} \right)$
$y = {{\rm{e}}^x}\lambda $, ${\ \ \ }$ $x=-1$
So, $y = {{\rm{e}}^{ - 1}}\lambda $.
Substituting $x=-1$ and $y = {{\rm{e}}^{ - 1}}\lambda $ in the constraint gives
${{\rm{e}}^{ - 1}} - {{\rm{e}}^{ - 1}}\lambda = 0$
So, $\lambda = 1$.
Using $x=-1$, $\lambda = 1$ and $y = {{\rm{e}}^{ - 1}}\lambda $, we obtain the critical point at $\left( { - 1,{{\rm{e}}^{ - 1}}} \right)$.
Thus, the critical value is $f\left( { - 1,{{\rm{e}}^{ - 1}}} \right) = - {{\rm{e}}^{ - 1}}$.
Next, we verify that $f\left( { - 1,{{\rm{e}}^{ - 1}}} \right) = - {{\rm{e}}^{ - 1}}$ is a minimum:
Substituting $y = {{\rm{e}}^x}$ in $f$ gives a function of single variable $h\left( x \right) = f\left( {x,{{\rm{e}}^x}} \right) = x{{\rm{e}}^x}$.
The derivatives of $h$ are
$h'\left( x \right) = {{\rm{e}}^x} + x{{\rm{e}}^x}$
$h{\rm{''}}\left( x \right) = {{\rm{e}}^x} + {{\rm{e}}^x} + x{{\rm{e}}^x} = \left( {2 + x} \right){{\rm{e}}^x}$
At the critical point $x=-1$, we have $h{\rm{''}}\left( { - 1} \right) = {{\rm{e}}^{ - 1}} > 0$.
Since $h{\rm{''}}\left( { - 1} \right) > 0$, we conclude that $f\left( { - 1,{{\rm{e}}^{ - 1}}} \right) = - {{\rm{e}}^{ - 1}}$ is a minimum.
Hence, the point $\left( {a,b} \right)$ on the graph of $y = {{\rm{e}}^x}$ where the value $a b$ is as small as possible occurs at the point $\left( { - 1,{{\rm{e}}^{ - 1}}} \right)$.
