What is the area of the largest rectangle in the first quadrant with two sides on the axes and one vertex on the curve y=e−x.
A=xy=xe−x=xex
If we want to maximize the area we set A′(x)=0 and solve for the dimensions of the box. So,
If A=xex, then by Quotient RuleA′(x)=ex(1)−(x)ex(ex)2A′(x)=ex−xex(ex)2If A′(x)=00=ex(1−x)
We have,
1−x=0 and ex=0 (we don't have solution for this)
So,
x=1
Therefore,
the largest area is A=xex=1e1=1e Square units
No comments:
Post a Comment