The hypotenuse of a right triangle has one end at the origin and one end on the curve y = x 2 e −3x , 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. At what x-value does it occur?

Answer:At x = 1 and maximum area  = 0.0499Step-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 =  xHeight 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 = 0For 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 = 1Maximum area, [tex]A=\dfrac{1}{e^3}\approx 0.0499[/tex]