area=1/2 y x
but y=above, so
area=1/2 x^5(e^-7x)
darea/dx=5/2 x^4(e^-7x)-7/2 x^5(e^-7x)=0
divide by x^4 e^-7x
0=5/2 -7/2x
x=5/7
y=you do it, then solve for Area.
check my work.
The hypotenuse of a right triangle has one end at the origin and one end on the curve y = (x^4)(e^-7x), with x > 0. One of the other two sides is on the x-axis, the other side is parallel to the y-axis. Find the maximum area of such a triangle.
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