A line in the first quadrant is tangent to the graph of y=1/x^2. How can we minimize the area between the line and the axes?
the line touching the graph at x=a has slope -2/a^3
The line through (a,1/a^2) with slope -2/a^3 is
y = -2/a^3 (x-a) + 1/a^2
This line crosses the axes at (0,3/a^2) and (3a/2,0)
So, the area of the triangle formed by the line and the axes is
A(a) = 1/2 * 3/a^2 * 3a/2 = 9/(4a)
Now we have us a problem. dA/da is never 0. The area is infinite at a=0, and goes to zero at a = oo
To minimize the area between the line and the axes, we need to find the line that is tangent to the graph of y = 1/x^2 in the first quadrant.
Let's start by finding the derivative of y = 1/x^2 to find the slope of the tangent line.
Taking the derivative of y = 1/x^2, we get:
dy/dx = -2/x^3
To find the slope of the tangent line at a particular point on the graph, we substitute the x-coordinate of that point into the derivative.
Let's assume that the desired tangent line intersects the graph at (a, 1/a^2), where a is a positive constant.
Substituting a into the derivative, we get:
dy/dx = -2/a^3
Now, we have the slope of the tangent line at the point (a, 1/a^2), which is -2/a^3.
Using point-slope form, we can write the equation of the tangent line:
y - 1/a^2 = (-2/a^3)(x - a)
Now, we need to find the point where this tangent line intersects the x-axis.
Since the y-coordinate of the x-axis is 0, we substitute y = 0 into the equation of the tangent line:
0 - 1/a^2 = (-2/a^3)(x - a)
Simplifying the equation, we get:
1/a^2 = (-2/a^3)(x - a)
To find the x-coordinate of the point of intersection, we solve this equation for x:
1/a^2 = -2(x - a)/a^3
Simplifying further:
1 = -2(x - a)/a
1 = -2x/a + 2
2x/a = 1
x = a/2
So, the point where the tangent line intersects the x-axis is (a/2, 0).
Now, let's find the point where the tangent line intersects the y-axis.
Since the x-coordinate of the y-axis is 0, we substitute x = 0 into the equation of the tangent line:
y - 1/a^2 = (-2/a^3)(0 - a)
Simplifying:
y - 1/a^2 = 2/a^2
y = 3/a^2
So, the point where the tangent line intersects the y-axis is (0, 3/a^2).
Now, we have the two points where the tangent line intersects the axes: (a/2, 0) and (0, 3/a^2).
To find the area between the line and the axes, we calculate the area of the triangle formed by these three points.
The base of the triangle is the x-coordinate difference between the two intersection points: a/2 - 0 = a/2.
The height of the triangle is the y-coordinate difference between the two intersection points: 3/a^2 - 0 = 3/a^2.
Therefore, the area of the triangle is:
Area = (1/2) * (base) * (height)
Area = (1/2) * (a/2) * (3/a^2)
Area = 3/4a
Now, we can minimize the area by finding the minimum value of the expression 3/4a.
To minimize the area, we consider the limiting cases as a approaches infinity.
As a approaches infinity, the area approaches 0.
Therefore, the minimum area between the line and the axes is 0, and it occurs when the tangent line is parallel to the x-axis, which happens as a approaches infinity.