Fix the vertices to have [tex]x[/tex]-coordinates [tex]x[/tex] and [tex]-x[/tex]. Then the rectangle has base [tex]x-(-x)=2x[/tex] and its height is [tex]e^{-3x^2}[/tex], the value of [tex]y[/tex] at these vertices' [tex]x[/tex]-coordinates.
Then the area of the rectangle is
[tex]A(x)=2xe^{-3x^2}[/tex]
Take the derivative:
[tex]A'(x)=2e^{-3x^2}(1-6x^2)[/tex]
We have [tex]A'(x)=0[/tex] when [tex]x=\dfrac1{\sqrt6}[/tex], at which point we get an area of
[tex]A\left(\dfrac1{\sqrt6}\right)e^{-3(1/\sqrt6)^2}=\sqrt{\dfrac2{3e}}\approx1.346[/tex]
The area of the rectangle is the products of its dimensions
The decimal approximation of the maximum area is 0.495
The given parameter is:
[tex]\mathbf{y = e^{-3x^2}}[/tex]
One side is on the x-axis.
So, the other dimension (d) is:
[tex]\mathbf{d = x -(-x)}[/tex]
[tex]\mathbf{d = x +x}[/tex]
[tex]\mathbf{d = 2x}[/tex]
The area of the rectangle is
[tex]\mathbf{A = d \times y}[/tex]
This gives
[tex]\mathbf{A =2x \times e^{-3x^2}}[/tex]
[tex]\mathbf{A =2xe^{-3x^2}}[/tex]
Differentiate using product rule:
[tex]\mathbf{A' =2e^{-3x^2} - 12x^2e^{-3x^2}}}[/tex]
Factorize
[tex]\mathbf{A' =2e^{-3x^2}(1 - 6x^2)}}[/tex]
Set to 0
[tex]\mathbf{2e^{-3x^2}(1 - 6x^2)} = 0}[/tex]
Solve for x
[tex]\mathbf{1 - 6x^2 = 0}[/tex]
Collect like terms
[tex]\mathbf{6x^2 = 1}[/tex]
Divide both sides by 6
[tex]\mathbf{x^2 = \frac 16}[/tex]
Take square roots
[tex]\mathbf{x =\sqrt{ \frac 16}}[/tex]
Substitute [tex]\mathbf{x =\sqrt{ \frac 16}}[/tex] in [tex]\mathbf{A =2x \times e^{-3x^2}}[/tex]
[tex]\mathbf{A =2 \times \sqrt{1/6} \times e^{-3(\sqrt{1/6})^2}}[/tex]
[tex]\mathbf{A =2 \times \frac{1}{\sqrt 6} \times e^{-3 \times 1/6}}[/tex]
[tex]\mathbf{A =2 \times \frac{1}{\sqrt 6} \times e^{-1/2}}[/tex]
[tex]\mathbf{A =0.495}[/tex]
Hence, the decimal approximation of the maximum area is 0.495
Read more about areas at:
https://brainly.com/question/12499175
