I asked for help on a question yesterday and I got really good help and I was wondering if someone could help with this problem too.
This is the question:
Find an equation of the line through the point (1,3) that cuts off the least area from the first quadrant.
My work on the problem so far:
I know that it will be in the first quadrant and that the result will form a right triangle.
I have written the equation y=-mx+b because I do know that the line will be decreasing.
Besides that, I do not know what to do.
Can someone help me please?
If you take the equation for the line to be:
y=-mx+b (1)
then you have to require that if x = 1, y = 3. So, you insert these figures in Eq. (1) and you get:
3 = -m + b (2)
This equation does not fix m and b but relates the two. This is what you would expect, because you can draw a line through the point (1,3) in an infinite number of ways. You then calculate where the line will intersect the x-axis. At that point y=0, so:
-mx + b =0 -->
x = b/m
And the line intersects the Y-axis when x = 0, so the y-coordinate at that point is b.
This maens that the right traingle as its three corners at:
(0,0), (0,b) and (b/m,0)
The area of the triangle is thus:
A = 1/2 b^2/m (3)
If you use Eq. (2) to express b in terms of m:
b = 3 + m and insert that in (3), you find:
A = 1/2 (3 + m)^2/m =
1/2 (9/m + 6 + m)
If you equate the derivative of this to zero you find:
m = 3.
In order to find the equation of the line that cuts off the least area from the first quadrant and passes through the point (1,3), you can start by assuming the equation has the form y = -mx + b, where m represents the slope of the line and b represents the y-intercept.
Since the line passes through the point (1,3), substitute these coordinates into the equation to get:
3 = -m(1) + b
Simplify this equation to:
3 = -m + b
This equation does not uniquely determine the values of m and b, but it establishes a relationship between them. To find where the line intersects the x-axis (where y = 0), set y = 0 in the equation:
0 = -mx + b
Solve for x to find:
x = b/m
This means that the line intersects the x-axis at the point (b/m, 0). Similarly, to find where the line intersects the y-axis (where x = 0), we can plug x = 0 into the equation:
y = -m(0) + b
Solve for y to find:
y = b
This means that the line intersects the y-axis at the point (0, b).
Now, consider the right triangle formed by the line, the x-axis, and the y-axis. The three corners of this triangle are (0,0), (0,b), and (b/m,0).
The area of this triangle can be calculated using the formula for the area of a triangle: A = (1/2) * base * height. In this case, the base of the triangle is b/m and the height is b. Substituting these values into the formula, we get:
A = (1/2) * (b/m) * b
Simplifying this expression, we have:
A = (1/2) * (b^2/m)
Now, substitute the relationship between m and b from equation (2) into equation (3) to get:
A = (1/2) * ((3 + m)^2/m)
Expanding and simplifying the expression, we have:
A = (1/2) * (9/m + 6 + m)
To find the value of m that minimizes the area A, we can take the derivative of A with respect to m and set it equal to zero. Find the derivative:
dA/dm = 9/(2m^2) + 1
Set the derivative equal to zero and solve for m:
9/(2m^2) + 1 = 0
Simplify the equation to:
9 + 2m^2 = 0
Now, solve for m:
2m^2 = -9
m^2 = -9/2
m = √(-9/2)
As the square root of a negative number is not real, it means there is no real value of m that minimizes the area A. This suggests that there is no line through the point (1,3) that cuts off the least area from the first quadrant.