The hypotenuse of a right triangle has one end at the origin and one end on the curve y+x^2e^(-3x), with x > or = 0. One of the other two sides is on the x-asis, the other side is parallel to the y-axis. Find the maximum area of such a triangle. At what x-value does it occur?
You appear to have a typo,
I will assume your curve is y = x^2e^(-3x) or y = x^2/e^(3x)
let the point of contact of the hypotenuse be (x,y) on the curve.
Then the right angle will be at (x,0)
and the
Area = xy/2
= (1/2)(x)x^2e^(-3x)
2A = (x^3)(e^(-3x))
2dA/dx = x^3(-3e^(-3x)) + 3x^2(e^(-3x))
= -3x^2(e^(-3x))[x - 1}
= 0 for a max of A
3x^2 = 0 --->x=0, little sense, since no triangle
or
e^(-3x) = 0 ---> no solution
or
x = 1, yeahh!
if x = 1 then
A = (1/2)1^3(e^-3) = .0249
( I tested for the area with x = .99 and x = 1.01 and they were both smaller than .0249 by a "smidgeon")
thank you and yes i had atypo i apologized it was y = x^2e^(-3x
To find the maximum area of the right triangle, we need to determine the length of the hypotenuse and one of the other two sides.
Let's start by finding the length of the hypotenuse. The hypotenuse of a right triangle is the longest side and is opposite the right angle. It can be found using the distance formula in Cartesian coordinates.
The equation of the curve given is y + x^2e^(-3x). To find the length of the hypotenuse, we need to find the distance between the origin (0,0) and a point on the curve.
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, (x1, y1) = (0, 0) and (x2, y2) = (x, y + x^2e^(-3x)). So the distance between the origin and a point on the curve is:
d = sqrt((x - 0)^2 + (y + x^2e^(-3x) - 0)^2)
= sqrt(x^2 + (y + x^2e^(-3x))^2)
Next, we need to find the length of one of the other two sides. One side is on the x-axis, so its length is simply x.
Now, we can proceed to find the area of the right triangle, which is given by:
Area = 1/2 * base * height
In this case, the base is x and the height is the length of the hypotenuse (d). So, the area can be written as:
Area = 1/2 * x * √(x^2 + (y + x^2e^(-3x))^2)
To find the maximum area, we need to find where the derivative of the area with respect to x is equal to zero. So, let's differentiate the area equation:
d(Area)/dx = 1/2 * (√(x^2 + (y + x^2e^(-3x))^2)) + (1/2 * x * (x^2 + (y + x^2e^(-3x))^2)^(-1/2) * (2x + (2x * e^(-3x) - 3x^2e^(-3x))))
= (√(x^2 + (y + x^2e^(-3x))^2) + x * (2x + (2x * e^(-3x) - 3x^2e^(-3x))) / (2 * √(x^2 + (y + x^2e^(-3x))^2)))
Setting this derivative equal to zero and solving for x will give us the x-value where the maximum area occurs.
Solving d(Area)/dx = 0 requires finding the root(s) of the equation.
By solving this equation, you will obtain the x-value(s) where the maximum area occurs. Once you have the x-value(s), you can substitute it back into the area equation to find the maximum area of the triangle.
Your curve
y+x^2 e^(-3x)
what is that equal to ? A curve has to follow an equation.
If you mean the curve
y=x^2 e^(-ex)
then the area is yx/2 which is equal to
1/2 x^3 e^(-3x). Right?
dArea/dx= 3x^2e^(-3x)-3x^3e^(-3x)=0
solve for x.
x=1 check that I did it in my head. Mornings is not a great time for me to try that.