Answer to Question #331743 in Calculus for Kai

Question

a rectangle is inscribed under the curve y=2^-x with its base along the positive x-axis. Find the dimensions of the rectangle with the largest area

Answer 1

Define:

$x$ - length of the rectangle $(x>0)$

$2^{-x}$ - height of the rectangle $(x>0)$

Area of the rectangle:

$S(x)=x\cdot2^{-x}$

$S’(x)=2^{-x}-x\cdot2^{-x}\cdot\ln2=0$

$x=\frac {1}{\ln2}$

$\displaystyle S’(\frac{1}{2\ln2})=2^{-\frac{1}{2\ln2}}(1-\frac{1}{2\ln2}\cdot\ln2)>0$

$\displaystyle S’(\frac{2}{\ln2})=2^{-\frac{2}{\ln2}}(1-\frac{2}{\ln2}\cdot\ln2)<0$

So $x=\frac {1}{\ln2}$ corresponds to maximum of the function $S(x)$ .

Answer:

$\frac {1}{\ln2}$ - length of the rectangle

$2^{-\frac {1}{\ln2}}$ - height of the rectangle