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

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

Step-by-step explanation:

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

(Eq. 1)

Where:

- Independent variable, dimensionless.

- Dependent variable, dimensionless.

- Slope, dimensionless.

- y-Intercept, dimensionless.

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

x-Intercept

y-Intercept

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

(Eq. 2)

If we know that , then the equation of the line is reduced into this:

(Eq. 3)

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

(Eq. 4)

The first and second derivatives are, respectively:

(Eq. 5)

(Eq. 6)

Then the first derivative is equalized to zero and solved:

And the second derivative is evaluated at critical value:

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

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

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