Answer:
[tex]A\approx 0.55[/tex]
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
[tex]f'(x)=0[/tex]
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
[tex]f(x)=e^{-x^2}[/tex]
The area of the rectangle is the product of the base by the height, so
[tex]A=xe^{-x^2}[/tex]
Let's find the first derivative
[tex]A'=e^{-x^2}-2x^2e^{-x^2}[/tex]
[tex]A'=e^{-x^2}(1-2x^2)[/tex]
Setting A'=0
[tex]e^{-x^2}(1-2x^2)=0[/tex]
[tex](1-2x^2)=0[/tex]
Solving for x
[tex]\displaystyle x=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}=0.707[/tex]
Let's compute the second derivative
[tex]A''=-2xe^{-x^2}(1-2x^2)+e^{-x^2}(-4x)[/tex]
[tex]A''=e^{-x^2}(4x^3-6x)[/tex]
Factoring
[tex]A''=2xe^{-x^2}(2x^2-3)[/tex]
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
[tex]2x^2-3[/tex]
Since [tex]x^2=1/2[/tex] the expression is negative, thus
A''(x)<0 and the critical point is a maximum
The maximum area is
[tex]A=\frac{\sqrt{2}}{2}e^{-(\frac{\sqrt{2}}{2})^2}[/tex]
[tex]\boxed{A\approx 0.55}[/tex]
