Question

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 1

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, $y=x^2e^{-3x}$  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, $A=\dfrac{1}{2}xy$

$A=\dfrac{1}{2}x^3e^{-3x}$

For maximum/minimum value of area.

$\dfrac{dA}{dx}=\dfrac{3}{2}x^2e^{-3x}-\dfrac{3}{2}x^3e^{-3x}$

Now, find critical point, $\dfrac{dA}{dx}=0$

$\dfrac{3}{2}x^2e^{-3x}-\dfrac{3}{2}x^3e^{-3x}=0$

$\dfrac{3}{2}x^2e^{-3x}(1-x)=0$

x =0,1

For x = 0, y = 0

For x = 1, $y=e^{-3}$

using double derivative test:-

$\dfrac{d^2A}{dx^2}=\dfrac{6}{2}xe^{-3x}-\dfrac{9}{2}x^2e^{-3x}-\dfrac{9}{2}x^2e^{-3x}-\dfrac{9}{2}x^3e^{-3x}$

At x= 0 , $\dfrac{d^2A}{dx^2}=0$

Neither maximum nor minimum

At x = 1, $\dfrac{d^2A}{dx^2}=-0.14$

Maximum area at x = 1

The maximum area of right triangle at x = 1

Maximum area, $A=\dfrac{1}{e^3}\approx 0.0499$