A rectangle has one side on the x-axis, one side on the y-axis, one vertex at the orgin ans one on the curve y = e^-3x , for x>(equal) 0. find the maximum area (using calculus)
a = xy = xe^(-3x)
da/dx = (1-3x)e^(-3x)
so, where is da/dx = 0?
is the answer 3?
no.
e^(-3x) is never zero, so where is (1-3x)=0?
x = 1/3
so, a = (1/3)e^(-3*1/3) = 1/(3e)
don't forget your Algebra I now that you're in Calculus!
To find the maximum area of a rectangle, we need to find the dimensions (length and width) that maximize the area. In this case, we are given that one side of the rectangle lies on the x-axis, another side lies on the y-axis, and one vertex is at the origin.
Let's assume the length of the rectangle is on the x-axis and is given by "x". The width of the rectangle, therefore, will be given by the value of y on the curve \(y = e^{-3x}\) at that particular x-axis value.
To find the area of the rectangle, we multiply the length and width: \(A = xy\). We can rewrite this expression as \(A = x \cdot e^{-3x}\).
Now, to find the maximum area, we need to differentiate the area function with respect to x and set the derivative equal to zero. This will give us the critical points.
\(A' = \frac{d}{dx}(x \cdot e^{-3x})\)
Applying the product rule:
\(A' = 1 \cdot e^{-3x} + x \cdot \frac{d}{dx}(e^{-3x})\)
\(A' = e^{-3x} - 3x \cdot e^{-3x}\)
Setting \(A'\) equal to zero:
\(e^{-3x} - 3x \cdot e^{-3x} = 0\)
Now, we can solve this equation for x to find the critical points.
Factor out \(e^{-3x}\):
\(e^{-3x}(1 - 3x) = 0\)
From this equation, we have two possibilities:
1) \(e^{-3x} = 0\) - This equation has no solutions since \(e^{-3x}\) is always positive.
2) \(1 - 3x = 0\) - Solve this equation to find the value of x:
\(1 - 3x = 0\)
\(3x = 1\)
\(x = \frac{1}{3}\)
We have found a critical point x = 1/3.
Now, we need to check whether this point gives us a maximum area or not. To do that, we need to determine the second derivative of the area function:
\(A'' = \frac{d^2}{dx^2}(e^{-3x} - 3x \cdot e^{-3x})\)
Using the derivative rules and simplifying, we get:
\(A'' = 9x \cdot e^{-3x} - 15 \cdot e^{-3x}\)
Now, evaluate the second derivative at x = 1/3 to determine the concavity:
\(A''(1/3) = 9(1/3) \cdot e^{-(3(1/3))} - 15 \cdot e^{-(3(1/3))}\)
\(A''(1/3) = 3e^{-1} - 15e^{-1}\)
\(A''(1/3) = -12e^{-1}\)
Since the second derivative is negative at x = 1/3, the concavity is concave down.
Therefore, x = 1/3 is a maximum point.
Now, substitute x = 1/3 back into \(A = x \cdot e^{-3x}\) to find the maximum area:
\(A = \frac{1}{3} \cdot e^{-3(\frac{1}{3})}\)
Calculate this expression to find the maximum area.
Thus, using calculus, we can determine that the maximum area of the rectangle is given by the calculated value of A.