Please help out. Thank you.
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.
No one has answered this question yet.
Yes it is the right equation. Good job!
is the max. area 3.6945?
thanks.
Yes, that's correct too.
In exams and in general, if the number is a "round" number like e⊃2/2, I would give the answer as e²/2=3.6945(approx.)
because the latter is an approximation. This will give the exact value as the answer.
To find the maximum area of the right triangle, we need to first find the value of x that maximizes the area, and then substitute that value back into the equation to find the maximum area.
Let's start by finding the equation for the height of the triangle, y. We are given that the vertex of the triangle touches the graph of y=e^x. Since the right triangle lies in the first quadrant, the height of the triangle must be positive.
To find the y-coordinate of the vertex of the triangle, we substitute x=3 into the equation y=e^x:
y = e^3
Therefore, the equation of the height of the triangle is y = e^3.
Next, we need to find the equation for the base of the triangle, x. We are given that one leg of the right triangle lies on the x-axis, and the vertex on the x-axis is at the point (3,0).
Therefore, the equation of the base of the triangle is x = 3.
Now we can find the maximum area of the triangle by substituting the values of x and y into the formula for the area. The area of a triangle is given by the formula A = (base * height) / 2.
Substituting the values, we have:
A = (3 * e^3) / 2
So, the equation for the maximum area of the triangle is A = (3 * e^3) / 2.
To justify this answer, we need to show that this is indeed the maximum area. We can do this by taking the derivative of the area equation with respect to x, setting it equal to zero, and checking that the second derivative is negative at that point. However, since this may involve calculus, I will not explain the step-by-step process here unless requested.