Given:
- A straight line passing through point P(1, 4) with a negative slope.
- The line cuts the coordinate axes at points A and B.

To find:
The area of the triangle OAB when OA OB is minimum.

Solution:
Let the equation of the straight line passing through point P(1, 4) be y = mx + c, where m is the slope and c is the y-intercept.

Finding the equation of the line:
Since the line passes through point P(1, 4), we can substitute the values of x and y into the equation to find the value of c.
4 = m(1) + c
4 = m + c
c = 4 - m

Therefore, the equation of the line is y = mx + (4 - m).

Finding the coordinates of points A and B:
To find the coordinates of point A, we substitute x = 0 into the equation of the line.
y = m(0) + (4 - m)
y = 4 - m

Therefore, the coordinates of point A are (0, 4 - m).

To find the coordinates of point B, we substitute y = 0 into the equation of the line.
0 = mx + (4 - m)
mx = m - 4
x = (m - 4)/m

Therefore, the coordinates of point B are ((m - 4)/m, 0).

Finding the area of triangle OAB:
The coordinates of point O are (0, 0).

Using the distance formula, we can find the lengths of the sides OA and OB.
OA = sqrt((0 - 0)^2 + (4 - m)^2) = sqrt((4 - m)^2) = 4 - m
OB = sqrt((0 - (m - 4)/m)^2 + (0 - 0)^2) = sqrt((m - 4)^2/m^2) = (m - 4)/m

The area of a triangle can be found using the formula:
Area = (1/2) * base * height

In this case, the base is OA and the height is OB.
Therefore, the area of triangle OAB is:
Area = (1/2) * (4 - m) * ((m - 4)/m)

Finding the minimum area:
To find the minimum area, we take the derivative of the area function with respect to m and set it equal to zero.
d(Area)/dm = -1/2 + 4/m - 1/2 = 0
4/m = 1
m = 4

Substituting m = 4 into the area function:
Area = (1/2) * (4 - 4) * ((4 - 4)/4) = 0

Therefore, the minimum area of triangle OAB is 0.

Conclusion:
The area of the triangle OAB when OA OB is minimum is 0. Hence, the correct answer is option A.