Answer:
The equation of the line through the point (2, 5) that cuts off the least area from the first quadrant is [tex]y = -2.321\cdot x +9.642[/tex].
Step-by-step explanation:
From Analytical Geometry we get that the equation of the line is represented by:
[tex]y = m\cdot x + b[/tex] (Eq. 1)
Where:
[tex]x[/tex] - Independent variable, dimensionless.
[tex]y[/tex] - Dependent variable, dimensionless.
[tex]m[/tex] - Slope, dimensionless.
[tex]b[/tex] - y-Intercept, dimensionless.
From (Eq. 1) we find that intercepts are:
x-Intercept
[tex]x_{i} = -\frac{b}{m}[/tex]
y-Intercept
[tex]y_{i} = b[/tex]
And the area done by the line in the first quadrant is:
[tex]A = \frac{1}{2}\cdot x_{i}\cdot y_{i}[/tex]
[tex]A = \frac{1}{2}\cdot \left(-\frac{b}{m} \right) \cdot b[/tex]
[tex]A = -\frac{b^{2}}{2\cdot m}[/tex] (Eq. 2)
If we know that [tex](x, y) = (2,5)[/tex], then the equation of the line is reduced into this:
[tex]2\cdot m + b = 5[/tex]
[tex]b = 5-2\cdot m[/tex] (Eq. 3)
And we apply (Eq. 3) in (Eq. 2):
[tex]A = -\frac{(5-2\cdot m)^{2}}{2\cdot m}[/tex]
[tex]A = -\frac{25-20\cdot m +4\cdot m^{2}}{2\cdot m}[/tex]
[tex]A = -\frac{25}{2}\cdot m^{-2}-10+2\cdot m[/tex] (Eq. 4)
The first and second derivatives are, respectively:
[tex]A' = 25\cdot m^{-3}+2[/tex] (Eq. 5)
[tex]A'' = -75\cdot m^{-4}[/tex] (Eq. 6)
Then the first derivative is equalized to zero and solved:
[tex]25\cdot m^{-3}+2=0[/tex]
[tex]25\cdot m^{-3} = -2[/tex]
[tex]25 = -2\cdot m^{3}[/tex]
[tex]m^{3} = -\frac{25}{2}[/tex]
[tex]m = -2.321[/tex]
And the second derivative is evaluated at critical value:
[tex]A'' = -75\cdot (-2.321)^{-4}[/tex]
[tex]A'' = -2.584[/tex]
The critical value is associated with a positive area. Then, the y-intercept is: ([tex]m = -2.321[/tex])
[tex]b = 5-2\cdot (-2.321)[/tex]
[tex]b = 9.642[/tex]
Therefore, the equation of the line through the point (2, 5) that cuts off the least area from the first quadrant is [tex]y = -2.321\cdot x +9.642[/tex].
