A right triangle has one leg on the x-axis. The vertex at the right end of that leg is at the point (3,0). The other vertex touches the graph os y=e^x. the entire triangle is to lie in the first quadrant. Find the maximum area of this triangle.Justify your answer.
My work:
A= (3-x)y/2
y=e^x
A=(3-x)(e^x)/2 <-- is this the right equation?
Thanks.
Yes, you are on the right track. The equation A = (3 - x)(e^x)/2 represents the area of the right triangle.
To find the maximum area, we need to find the x-value that maximizes A. To do this, we can take the derivative of A with respect to x and set it equal to 0.
Let's differentiate A with respect to x:
A' = [(3 - x)(e^x)/2]'
= [(3 - x)' * (e^x) + (3 - x) * (e^x)']/2
= (-1 * (e^x) + (3 - x) * (e^x)')/2
= (-e^x + (3 - x)e^x')/2
= (-e^x + (3 - x)e^x)/2
Setting A' equal to 0 and solving for x:
(-e^x + (3 - x)e^x)/2 = 0
-e^x + (3 - x)e^x = 0
(3 - x)e^x = e^x
3 - x = 1
x = 2
Now, we need to check if this critical point is actually a maximum. We can do this by taking the second derivative of A and evaluating it at x = 2.
Let's find A'' by differentiating A':
A'' = (-1 * e^x + (-1)e^x + (3 - x)'e^x + (3 - x)e^x')/2
= (-2 * e^x + (-1)e^x' + (3 - x)' * e^x + (3 - x)e^x')/2
= (-2 * e^x - e^x' + (3 - x)' * e^x + (3 - x)e^x')/2
= (-2 * e^x - e^x' + (-1)e^x + (3 - x)e^x')/2
= (-2 * e^x - 2 * e^x - e^x + (3 - x)e^x')/2
= (-3 * e^x + (3 - x)e^x')/2
= (-3 * e^x + (3 - x)e^x)/2
Evaluating A'' at x = 2:
A''(2) = (-3 * e^2 + (3 - 2)e^2)/2
= (-3 * e^2 + e^2)/2
= (-2 * e^2)/2
= -e^2 < 0
Since A''(2) is negative, this confirms that x = 2 is a maximum point.
Therefore, the maximum area of the right triangle is given by:
A(2) = (3 - 2)(e^2)/2
= (1)(e^2)/2
= e^2/2
So, the maximum area of the triangle is e^2/2.
To find the maximum area of the triangle, we need to find a function that represents the area of the triangle in terms of one variable. Let's call the height of the triangle h and the base of the triangle x.
We know that one leg of the triangle is on the x-axis and its vertex is at (3,0), so the length of the base is 3 - x.
The other vertex of the triangle lies on the graph of y = e^x. Since this vertex touches the graph, the height of the triangle will be equal to the y-coordinate of the point on the graph at x.
So, the equation for the area A of the triangle becomes:
A = (1/2) * base * height
A = (1/2) * (3 - x) * e^x
To find the maximum area, we can differentiate this equation with respect to x and equate it to 0, and then solve for x.
Let's find the derivative of A with respect to x:
dA/dx = (1/2)(-1)e^x + (1 - x)e^x
Setting this derivative equal to 0, we have:
(1/2)(-1)e^x + (1 - x)e^x = 0
Simplifying this equation, we get:
(-1/2)e^x + e^x - xe^x = 0
Multiplying both sides by 2, we have:
- e^x + 2e^x - 2xe^x = 0
Rearranging terms, we get:
e^x (1 - 2 + 2x) = 0
So, either e^x = 0 (which is not possible) or 1 - 2 + 2x = 0.
Solving for x, we have:
2x = 1 - 2
2x = -1
x = -1/2
Since the triangle needs to lie in the first quadrant, we discard the negative value of x.
Therefore, x = -1/2 is not a valid solution for our problem.
So, we conclude that the maximum area of the triangle is obtained when x = 0.
When x = 0, the base of the triangle is 3 - 0 = 3.
The height of the triangle is e^0 = 1.
Therefore, the maximum area of the triangle is:
A = (1/2) * (3 - 0) * 1 = 3/2.
Hence, the maximum area of the triangle is 3/2.