Question

Calculus 1 question on the photo

Answer 1

Answer:

$A\approx 0.55$

Step-by-step explanation:

Optimizing With Derivatives

One of the most-used applications of derivatives is to maximize or minimize functions. We need to recall that if f(x) is a real function and f'(x) is the derivative of f, then we can find the critical points of f by setting

$f'(x)=0$

Then we must test the critical points in the second derivative f''(x) and if

f''(x) is positive, then x is a minimum

f''(x) is negative, then x is a maximum

The problem requires us to find the maximum area of the rectangle which base is x and height is f(x), where

$f(x)=e^{-x^2}$

The area of the rectangle is the product of the base by the height, so

$A=xe^{-x^2}$

Let's find the first derivative

$A'=e^{-x^2}-2x^2e^{-x^2}$

$A'=e^{-x^2}(1-2x^2)$

Setting A'=0

$e^{-x^2}(1-2x^2)=0$

$(1-2x^2)=0$

Solving for x

$\displaystyle x=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}=0.707$

Let's compute the second derivative

$A''=-2xe^{-x^2}(1-2x^2)+e^{-x^2}(-4x)$

$A''=e^{-x^2}(4x^3-6x)$

Factoring

$A''=2xe^{-x^2}(2x^2-3)$

Evaluating for the critical point we can see the first factor (2x) is positive. The exponential is always positive, we only need to find the sign of

$2x^2-3$

Since $x^2=1/2$ the expression is negative, thus

A''(x)<0 and the critical point is a maximum

The maximum area is

$A=\frac{\sqrt{2}}{2}e^{-(\frac{\sqrt{2}}{2})^2}$

$\boxed{A\approx 0.55}$