Question

The area of the largest rectangle that can be drawn with one side along the x-axis and two vertices on the curve of y=e^-x^2 is...

Answer 1

so length of base of rectangle is 2x
height of rectangle is y or e^(-x^2)

this gives an area function:

Answer 2

\[A = 2x e^{-x^2}\]
differentiate and set equal to zero to maximize area

Answer 3

ok so derivative is \[A'= 2e ^{-x ^{2}}\]

Answer 4

or do you have to take derivative of -x^2 also?

Answer 5

using product rule
\[\frac{dA}{dx} = 2 e^{-x^2} -4x^2 e^{-x^2} = 0\]

Answer 6

ok

Answer 7

so what so we do next?

Answer 8

wait do you understand product rule and how i got derivative?

Answer 9

yea, i just always forget to use it :/

Answer 10

f'(x)= a'b +ab'

Answer 11

yeah ok, next just solve for x

Answer 12

factor out the e^-x^2 part

Answer 13

so e^-x^2(2-4x^2)

Answer 14

=0

Answer 15

yes

e^-x^2 = 0 has no solution
solve
2-4x^2 = 0

Answer 16

x=sqrt 1/2

Answer 17

so base of rectangle is 1

Answer 18

hmm no , base = 2x ----> 2*sqrt(1/2)

Answer 19

whoops

Answer 20

:/ so base is 2rt1/2 and then height is e^-1/2?

Answer 21

you can rationalize sqrt(1/2) to sqrt2/2

Answer 22

yep so max Area = 2sqrt(1/2)*e^-1/2

or
sqrt(2)*e^-1/2

Answer 23

ok, thank you a lot! I understand the concept, but i was making some silly math errors.

Answer 24

yw :)