Answer:
At x = 1 and maximum area = 0.0499
Step-by-step explanation:
The hypotenuse of a right triangle has one end at the origin and other end on the curve, [tex]y=x^2e^{-3x}[/tex] with x ≥ 0.
One leg of right triangle is x-axis and another leg parallel to y-axis.
Length of base of right triangle = x
Height of right triangle = y
Area of right triangle, [tex]A=\dfrac{1}{2}xy[/tex]
[tex]A=\dfrac{1}{2}x^3e^{-3x}[/tex]
For maximum/minimum value of area.
[tex]\dfrac{dA}{dx}=\dfrac{3}{2}x^2e^{-3x}-\dfrac{3}{2}x^3e^{-3x}[/tex]
Now, find critical point, [tex]\dfrac{dA}{dx}=0[/tex]
[tex]\dfrac{3}{2}x^2e^{-3x}-\dfrac{3}{2}x^3e^{-3x}=0[/tex]
[tex]\dfrac{3}{2}x^2e^{-3x}(1-x)=0[/tex]
x =0,1
For x = 0, y = 0
For x = 1, [tex]y=e^{-3}[/tex]
using double derivative test:-
[tex]\dfrac{d^2A}{dx^2}=\dfrac{6}{2}xe^{-3x}-\dfrac{9}{2}x^2e^{-3x}-\dfrac{9}{2}x^2e^{-3x}-\dfrac{9}{2}x^3e^{-3x}[/tex]
At x= 0 , [tex]\dfrac{d^2A}{dx^2}=0[/tex]
Neither maximum nor minimum
At x = 1, [tex]\dfrac{d^2A}{dx^2}=-0.14<0[/tex]
Maximum area at x = 1
The maximum area of right triangle at x = 1
Maximum area, [tex]A=\dfrac{1}{e^3}\approx 0.0499[/tex]
