SOLUTION: Find the area of the largest rectangle that has two sides on the positive x-axis and the positive y-axis one vertex at the origin and one vertex on the curve y = e^(-x)
Explain
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-> SOLUTION: Find the area of the largest rectangle that has two sides on the positive x-axis and the positive y-axis one vertex at the origin and one vertex on the curve y = e^(-x)
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Question 999522: Find the area of the largest rectangle that has two sides on the positive x-axis and the positive y-axis one vertex at the origin and one vertex on the curve y = e^(-x)
Explain how you know the value you found is the maximum area.
Thank you Answer by Theo(13342) (Show Source):
x is the length of the rectangle.
e^-x is the width of the rectangle.
the area will be maximum when x = 1.
you can solve this graphically, or you can use calculus to find the max/min point.
the graphical solution is shown below:
the area is equal to x * e^(-x).
you graph that function and find the maximum point on the curve.
the maximum point on the curve is when x = 1 and y = e^-1 = .3678794412.
this is shown on the graph as (1,.3679).
at that point, the value of the x-coordinate is 1 and the value of the y-coordinate is .3679.
the graphing software i used automatically finds the maximum point.
here's the graph:
the red line is y = e^(-x).
the blue line is y = x * e^(-x).
the blue line is the graph of the area of the rectangle.
the area of a rectangle is length * width.
in the formula of y = x * e^(-x), the length is x and the width is e^(-x).
you could also have used calculus.
the max/min point of the equation is when the derivative of the equation is equal to 0.
the derivative of x * e^(-x) is equal to -(x-1)*e^(-x)
set that equal to 0 to get -(x-1)*e^(-x) = 0
remove the parentheses to get -x * e^(-x) + e^(-x) = 0
add x*(e^(-x) to both sides of this equation to get e^(-x) = x*e^(-x)
divide both sides of this equation by e^(-x) to get 1 = x
solve for x to get x = 1.
note that, in the equation of -(x-1)*e^(-x) = 0, i could also have divided both sides of the equation by -(x-1) to get e^(-x) = 0.
i would have then solved that by taking the natural log of both sides of the equation to get:
ln(e^(-x)) = ln(0).
since ln(0) is invalid, that would not have yielded a good result.
i then went to the alternate method of removing the parentheses, which led to the correct result.
even though dividing both sides of the equation by -(x-1) was a valid algebraic operation, it did not provide a satisfactory solution to this problem because of the restrictions on the value that you can take the natural log of. you can only take the natural log of a positive number.
i used the following online derivative calculator to find the derivative of x * e^(-x).
