A rectangle is drawn in the first quadrant so that it has two adjacent sides on the coordinate axes and one
vertex on the curve y = − ln x. Find the x-coordinate of the vertex for which the area of the rectangle is a
maximum.
Draw a diagram. The area is
A = xy = -x lnx
dA/dt = -lnx - 1
A has a max of 1/e when x = 1/e
Ah, the quest for the maximum area of a rectangle drawn in the first quadrant! Well, let's put on our mathematical clown shoes and dive right in!
To find the x-coordinate of the vertex that maximizes the area of the rectangle, we need to figure out the length of the sides. One side of the rectangle is along the x-axis, so its length is simply x. The other side of the rectangle is along the y-axis and passes through the point on the curve y = -ln(x). Since the y-coordinate is negative, we can take the absolute value to get the length of the side, which is |-ln(x)|.
Now the formula for the area of a rectangle is just length times width, so the area A(x) of our rectangle is A(x) = x * |-ln(x)|.
To find the maximum area, we need to find the critical points of A(x) by taking its derivative and setting it equal to zero. Let's get our clown calculus act going!
A'(x) = | -ln(x) | - x * ( -1 / x ) = | -ln(x) | + 1.
Setting A'(x) equal to zero, we get | -ln(x) | + 1 = 0. Now, let's break out our clown logic: absolute values only equal zero if what's inside them equals zero. So, -ln(x) = 0.
Solving -ln(x) = 0, we find x = 1. Ah, the magic number!
Now, we need to check if this x-coordinate of 1 actually maximizes the area. We can use the second derivative test for that. Taking the second derivative of A(x), we get A''(x) = 1 / x^2.
Since A''(1) = 1 / 1^2 = 1, which is positive, we can conclude that x = 1 indeed corresponds to the maximum area. Ta-da!
So, the x-coordinate of the vertex that maximizes the area of the rectangle is 1. Now, go forth and impress your friends with your mathematical clownery!
To find the x-coordinate of the vertex for which the area of the rectangle is a maximum, we need to maximize the area of the rectangle.
Let's assume that the rectangle is formed by the x-axis, y-axis, and a vertical line passing through a point (x, -ln(x)) on the curve y = -ln(x).
The length of the rectangle along the x-axis is 2x (since it has two adjacent sides on the coordinate axes).
The height of the rectangle is given by the difference in y-values of the point (x, -ln(x)) and the x-axis, which is -ln(x).
So, the area of the rectangle is given by A = 2x * (-ln(x)) = -2x * ln(x).
To find the maximum area, we need to find the value of x that maximizes the area function A = -2x * ln(x).
To do this, we'll take the derivative of A with respect to x and set it equal to zero:
dA/dx = -2ln(x) - 2 = 0
Solving this equation for x, we get:
ln(x) = -1
Using the property of logarithms, we can rewrite the equation as:
x = e^(-1)
Therefore, the x-coordinate of the vertex for which the area of the rectangle is a maximum is x = e^(-1).
Hope this helps!
To find the x-coordinate of the vertex for which the area of the rectangle is a maximum, we need to determine the dimensions of the rectangle.
Let's break down the problem step by step:
Step 1: Find the equation of the line representing the two adjacent sides of the rectangle.
Since the rectangle has two adjacent sides on the coordinate axes, we can determine the equation of the line by using the equation of a straight line, y = mx + c.
For the vertical side on the y-axis, the x-coordinate is 0, so the equation becomes: y = m(0) + c => y = c.
For the horizontal side on the x-axis, the y-coordinate is 0, so the equation becomes: 0 = mx + c => mx = -c => x = -c/m.
Now, we have two points on the line: (0, c) and (-c/m, 0). We can use these two points to find the equation of the line.
Using the slope formula (m = (y2 - y1)/(x2 - x1)), we get:
m = (0 - c)/(-c/m - 0)
m = -c/(-c/m)
m = -m
Simplifying further, we get m² = 1 => m = ±1.
So the equations for the two adjacent sides of the rectangle are:
1. y = c, where m = 1
2. x = -c, where m = -1
Step 2: Find the x-coordinate of the vertex on the curve y = -ln(x).
To find the x-coordinate of the vertex, we need to determine the maximum point on the curve. Since the curve is concave down, the maximum point occurs at the point where the derivative is equal to zero.
The derivative of y = -ln(x) is dy/dx = -1/x.
Setting dy/dx = 0, we get -1/x = 0 => x = 0.
So the x-coordinate of the maximum point on the curve is 0.
Step 3: Find the x-coordinate of the vertex for which the area of the rectangle is a maximum.
Since the rectangle has two adjacent sides on the coordinate axes and one vertex on the curve, the area of the rectangle is given by:
Area = length * width
The width of the rectangle is the difference between the y-coordinate at the vertex (0) and the y-coordinate on the curve (y = -ln(x)). So the width is -ln(x) - 0 = -ln(x).
The length of the rectangle is the difference between the x-coordinate at the vertex (0) and the x-coordinate on the line y = c (x = -c). So the length is -c - 0 = -c.
Therefore, the area of the rectangle is:
Area = -c * (-ln(x)) = c * ln(x)
To find the x-coordinate of the vertex for which the area of the rectangle is a maximum, we need to maximize the area function c * ln(x).
Since c is a constant, we only need to maximize ln(x).
To find the maximum value of ln(x), we set its derivative equal to zero:
d/dx ln(x) = 1/x = 0
Solving for x, we get x = 1.
Therefore, the x-coordinate of the vertex for which the area of the rectangle is a maximum is x = 1.