Question

Calculus 1 question on the photo

Answer 1

`A\approx 0.55`

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{(1)/(2)}=(√(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=(√(2))/(2)e^{-((√(2))/(2))^2}`

`\boxed{A\approx 0.55}`