# A rectangle is made having 1 side on the x-axis and the 2 upper corner points on the graph of y = 4e^(−2x^2)+3). Determine the max. possible area, Amax, of this. The rectangle will have width 2x and height `4e^(-2x^2+3)` . (Note that the given function is symmetric about the y-axis.)

The area will be `A=(2x)(4e^(-2x^2+3))=8xe^(-2x^2+3)`

To maximize this function we take the first derivative and set it equal to zero in order to find the critical points:

`(dA)/(dx)=8e^(-2x^2+3)+(8x)(-4x)e^(-2x^2+3)` using the...

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The rectangle will have width 2x and height `4e^(-2x^2+3)` . (Note that the given function is symmetric about the y-axis.)

The area will be `A=(2x)(4e^(-2x^2+3))=8xe^(-2x^2+3)`

To maximize this function we take the first derivative and set it equal to zero in order to find the critical points:

`(dA)/(dx)=8e^(-2x^2+3)+(8x)(-4x)e^(-2x^2+3)` using the product rule

`=e^(-2x^2+3)[8-32x^2]` factoring out the common factor.

Setting this equal to zero we get:

`e^(-2x^2+3)[8-32x^2]=0 ==>8-32x^2=0` since e to a power is positive.

`32x^2=8==>x=+-1/2`

Thus the width of the rectangle is 1 (2*1/2) and the height is `f(1/2)=4e^(-2(1/2)^2+3)~~48.73`

The maximal area is approximately 48.73 square units.

The graph:

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