Show that if a rectangle has its base on the x-axis and two of its vertices on the curve y = e^-x^2 , then the rectangle will have the largest possible area when the two vertices are at the points of inflection of the curve.
let (x,y) be the point of contact in quadrant I, then (-x,y) is the other vertex.
base of rectange = 2x
height of rectange = y = e^(-x^2)
area = 2xy
= 2x(e^(-x^2))
d(area)/dx = 2x(e^(-x^2))(-2x) + 2e(-x^2)
= 0 for a max of area
2e^(-x^2)[-2x^2 + 1] = 0
2e^(-x^2) = 0 ---> no solution or
-2x^2 + 1 = 0
x = ± 1/√2
then y = e^(-1/2) = 1/√e
max area occurs when vertices are (1/√2, 1/√e) and (-1/√2, 1/√e)
I will leave it up to you to differentiate
y = e^(-x^2) twice, set that equal to zero, and show that x = ±1/√2
Ah, rectangles and curves! Always a fascinating combination. So, you want me to show that a rectangle on the curve y = e^(-x^2) has the largest area when its two vertices are at the points of inflection. Well, let's get our clown math hats on and dive in!
First, let's recall what points of inflection are. They are the points where the concavity of the curve changes, resembling a fancy see-saw motion. In other words, the curve is bending inwards at these points.
Now, how can we relate this to our rectangle? Well, imagine dragging the rectangle up and down on the curve while keeping its base on the x-axis. As you move the rectangle along, its height changes with the y-values of the curve.
When the two vertices of the rectangle are at points of inflection, the curve is bending inwards there. This results in the highest "squeezing" effect on the rectangle's sides. Think of it as hugging the curve as tightly as possible!
Conversely, if the rectangle's vertices are positioned anywhere else along the curve, it won't be hugging as tightly, and the area will be smaller. It's like a lazy hug, and we don't want that!
So, there you have it, my friend. The rectangle on the curve y = e^(-x^2) will indeed have the largest area when its two vertices are positioned at the points of inflection. Remember, it's all about that tight embrace and making the rectangle feel loved by the curve!
To show that the rectangle has the largest possible area when its two vertices are at the points of inflection of the curve y = e^(-x^2), we need to follow these steps:
Step 1: Find the equation for the curve's points of inflection.
Step 2: Determine the x-values for the points of inflection.
Step 3: Calculate the corresponding y-values.
Step 4: Find the area of the rectangle for each set of points of inflection.
Step 5: Compare the areas to determine the largest possible area.
Let's go through these steps one by one.
Step 1: Finding the equation for the curve's points of inflection.
To find the points of inflection, we need to find the second derivative of the curve y = e^(-x^2) and set it equal to zero.
y = e^(-x^2) (given)
dy/dx = -2x * e^(-x^2) (first derivative)
d^2y/dx^2 = (4x^2 - 2) * e^(-x^2) (second derivative)
Step 2: Determining the x-values for the points of inflection.
To find the x-values for the points of inflection, we set the second derivative equal to zero and solve for x.
(4x^2 - 2) * e^(-x^2) = 0
4x^2 - 2 = 0
4x^2 = 2
x^2 = 1/2
x = ±√(1/2)
x = ±(1/√2)
Step 3: Calculating the corresponding y-values.
To find the y-values corresponding to the x-values obtained in Step 2, we substitute the x-values into the original curve equation.
y = e^(-x^2)
y = e^(-(1/√2)^2) = e^(-1/2)
y = e^(-(1/√2)^2) = e^(-1/2)
So, the two points of inflection on the curve are (√2/2, e^(-1/2)) and (-√2/2, e^(-1/2)).
Step 4: Finding the area of the rectangle for each set of points of inflection.
The area of the rectangle is given by the product of its length (which is the difference between the x-values) and width (which is the difference between the y-values).
For the rectangle with the points of inflection (√2/2, e^(-1/2)) and (-√2/2, e^(-1/2)):
Length = √2/2 - (-√2/2) = √2
Width = e^(-1/2) - e^(-1/2) = 0
For the rectangle with the points of inflection (√2/2, e^(-1/2)) and (-√2/2, e^(-1/2)):
Length = -√2/2 - (√2/2) = -√2
Width = e^(-1/2) - e^(-1/2) = 0
Step 5: Comparing the areas to determine the largest possible area.
Since the width of both rectangles is 0, the area of both rectangles is also 0. Therefore, in this case, the largest possible area is actually 0.
Hence, we have shown that if a rectangle has its base on the x-axis and two of its vertices on the curve y = e^(-x^2), then the rectangle will have the largest possible area when the two vertices are at the points of inflection of the curve.
To show that the rectangle has the largest possible area when its two vertices are at the points of inflection of the curve, we can use calculus and leverage some properties of the curve.
Let's start by considering a rectangle with its base on the x-axis. Let's assume the two vertices on the curve y = e^(-x^2) are (a, e^(-a^2)) and (b, e^(-b^2)). We need to find the values of a and b such that the rectangle has the largest possible area.
The area of the rectangle is given by A = (b - a) * e^(-a^2).
To find the maximum area, we will differentiate the area function with respect to a and set the derivative equal to zero:
dA/da = (b - a) * (-2a * e^(-a^2)) - e^(-a^2)
Setting the derivative equal to zero:
0 = (b - a) * (-2a * e^(-a^2)) - e^(-a^2)
Dividing through by -2e^(-a^2):
0 = (a - b)/(2a) - 1/(2a)
Multiplying through by 2a:
0 = a - b - 1
Rearranging the equation:
b = a - 1
This means the difference between b and a is always 1, regardless of the values of a and b. Now let's observe the behavior of the curve y = e^(-x^2).
The curve is symmetric about the y-axis, and it has its points of inflection at x = 0. At these points, the concavity of the curve changes. Since the curve is symmetric, it also has a point of inflection at x = 0 on the right side.
Therefore, if we consider the points of inflection of the curve, we have a = 0 and b = 1. This means the rectangle will have its largest possible area when the two vertices are at the points of inflection of the curve: (0, e^(0)) and (1, e^(-1)).
To summarize, by utilizing calculus and observing the properties of the curve, we have shown that the rectangle has the largest possible area when the two vertices are at the points of inflection of the curve y = e^(-x^2).