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 $g(x,y)=y-e^{8x}$

Apply Lagrange Multiplier as follows:

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

So, find the gradients both sides and equate as,

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

Equate both sides to obtain,

$$y=-\lambda8e^{8x}$$ $$x=\lambda$$

Substitute $\lambda=x$ into equation $y=-\lambda8e^{8x}$ to obtain $y=-8xe^{8x}$

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

$$-8xe^{8x}-e^{8x}=0$$

Solve the equation as follows:

$$e^{8x}(-8x-1)=0$$

$$-8x-1=0,e^{8x}\ne0$$

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

Next, plug $x=-\frac{1}{8}$ into relation $y=-8xe^{8x}$ to find the value of $y$ as,

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

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