Question

Use Lagrange multipliers to find the point (a,b) on the graph of y=e8x, where the value ab is as small as possible.

P=?

Answer 1

As we know, for all points that are on the graph function $y=e^{8x}$ are performed

$$\boxed{\forall (a,b) : b=e^{8a}}$$

Then we reformulate our problem: find the minimum of the function $f(a,b)=ab$ with the condition $e^{8a}=b$ . We introduce the Lagrange function and study it for extremum

$$L(a,b;\lambda)=ab+\lambda\cdot\left(b-e^{8a}\right)\longrightarrow\\[0.3cm] \left\{\begin{array}{l} \displaystyle\frac{\partial L}{\partial a}=b-8\lambda e^{8a}=0\\[0.4cm] \displaystyle\frac{\partial L}{\partial b}=a+\lambda=0\\[0.4cm] \displaystyle\frac{\partial L}{\partial\lambda}=b-e^{8a}=0 \end{array}\right.\longrightarrow\\[0.4cm] b=8\lambda e^{8a}\longrightarrow e^{8a}=\frac{b}{8\lambda}\longrightarrow\\[0.4cm] b-\frac{b}{8\lambda}=0\longrightarrow b\cdot\left(\frac{8\lambda-1}{8\lambda}\right)=0\longrightarrow\\[0.4cm] 8\lambda-1=0\longrightarrow\boxed{\lambda=\frac{1}{8}}\longrightarrow\\[0.4cm] \left\{\begin{array}{l} a=-\lambda=-\displaystyle\frac{1}{8}\\[0.4cm] b=8\lambda e^{8a}=8\cdot\displaystyle\frac{1}{8}\cdot e^{8\cdot\left(-\frac{1}{8}\right)}=e^{-1} \end{array}\right.\longrightarrow\\[0.4cm]$$

Conclusion,

$$\min\limits_{(a,b)\in\mathbb{R}^2}f(a,b)=f\left(-\frac{1}{8},e^{-1}\right)=-\frac{1}{8}\cdot e^{-1}\\[0.3cm] \boxed{\min\limits_{(a,b)\in\mathbb{R}^2}f(a,b)=-\frac{1}{8e}\approx -0.4598}$$

To check, let's look at the graph of the investigated function

ANSWER

$$\min\limits_{(a,b)\in\mathbb{R}^2}f(a,b)=-\frac{1}{8e}\approx -0.4598$$