Determining slope that cuts off least area So here is the question: 

Determine the slope of the line that passes through the point $(1,2)$ and that cuts off the least area from the first quadrant. 

I've thought about this question, and I could only come up with the slope as a negative value, since it is decreasing. I think $y=-x+3$ would work but I don't know how to prove that the area cut off is the smallest. 
 A: Not sure what your experience with maximizing/minimizing functions is, but it will generally require taking the first derivative of an expression.  So lets create an expression for the area cut off in terms of $m$, the slope of the line.
If we trace the line to the left from $(1,2)$ we can see that it will cross the y-axis at a point of $2-m$ and if we trace to the right, we can see the line will cross the x-axis at a point of $\frac{-2}{m}+1$.  This creates a triangle whose area is $$\frac{1}{2}(h)(b)=\frac{1}{2}(2-m)(\frac{-2}{m}+1)$$
If you take the first derivative of that and set it to zero, you should get the slope you are after.

A: After you have found the answer by the standard method (setting a derivative to zero, as explained in another answer), you might want to try the following intuitive visualization:
Plot the solution of the problem as a line through the point $(1,2).$
It cuts off a right triangle from the first quadrant.
Now consider another line with a slightly steeper (or shallower) slope through
the same point.
By using that line instead of the line you found by your calculations,
you would change the area of the triangle by removing a thin triangular sliver
extending from $(1,2)$ to one of the axes along the right triangle's hypotenuse,
and adding a thin triangular sliver extending from $(1,2)$ to the other axis
along the right  triangle's hypotenuse.
If one of the two triangular slivers is relatively much larger than the other, 
then you could reduce the area
cut off by your line either by choosing a the new line instead or by
choosing a line that is "rotated" in the opposite direction.
You have a solution only if the two triangular slivers are almost exactly the same
area (and the only reason we have to say "almost" is that if the areas were exactly
the same then it wouldn't matter which of the two lines you took as the solution;
but it happens that there really is only one solution to this problem).
A: There is a solution to this problem that does not involve calculus:
If we modify the equation in the solution of @turkeyhundt, by letting $A$ represent the area of the triangle, we get:
$$A=\frac{1}{2}(2-m)(\frac{-2}{m}+1)$$ Multiplying through by $2m$ and rearranging, we get this quadratic in $m$:$$m^2+(2A-4)m+4=0$$The discriminant of this quadratic is $D$:$$D=(2A-4)^2-16$$This discriminant must be nonnegative to permit real values of $m$. The smallest positive $A$ that produces a non-negative discriminant is $A=4$;  substituting this in the quadratic gives the required value of $m$.
