Question

Find the equation of the line through the point (2, 5) that cuts off the least area from the first quadrant.

Answer 1

The equation of the line through the point (2, 5) that cuts off the least area from the first quadrant is
`y = -2.321\cdot x +9.642`.

Explanation:

From Analytical Geometry we get that the equation of the line is represented by:

`y = m\cdot x + b` (Eq. 1)

Where:

`x` - Independent variable, dimensionless.

`y` - Dependent variable, dimensionless.

`m` - Slope, dimensionless.

`b` - y-Intercept, dimensionless.

From (Eq. 1) we find that intercepts are:

x-Intercept

`x_(i) = -(b)/(m)`

y-Intercept

`y_(i) = b`

And the area done by the line in the first quadrant is:

`A = (1)/(2)\cdot x_(i)\cdot y_(i)`

`A = (1)/(2)\cdot \left(-(b)/(m) \right) \cdot b`

`A = -(b^(2))/(2\cdot m)` (Eq. 2)

If we know that
`(x, y) = (2,5)`, then the equation of the line is reduced into this:

`2\cdot m + b = 5`

`b = 5-2\cdot m` (Eq. 3)

And we apply (Eq. 3) in (Eq. 2):

`A = -((5-2\cdot m)^(2))/(2\cdot m)`

`A = -(25-20\cdot m +4\cdot m^(2))/(2\cdot m)`

`A = -(25)/(2)\cdot m^(-2)-10+2\cdot m` (Eq. 4)

The first and second derivatives are, respectively:

`A' = 25\cdot m^(-3)+2` (Eq. 5)

`A'' = -75\cdot m^(-4)` (Eq. 6)

Then the first derivative is equalized to zero and solved:

`25\cdot m^(-3)+2=0`

`25\cdot m^(-3) = -2`

`25 = -2\cdot m^(3)`

`m^(3) = -(25)/(2)`

`m = -2.321`

And the second derivative is evaluated at critical value:

`A'' = -75\cdot (-2.321)^(-4)`

`A'' = -2.584`

The critical value is associated with a positive area. Then, the y-intercept is: (
`m = -2.321`)

`b = 5-2\cdot (-2.321)`

`b = 9.642`

Therefore, the equation of the line through the point (2, 5) that cuts off the least area from the first quadrant is
`y = -2.321\cdot x +9.642`.