1. ## intercept

a line pass through (1,4).determine the least value of sum of the intercept on the coordinate axes.
i am getting 3,but book answer is 9.who is wrong.(use calculus only)

2. If you are considering intercepts for x, y > 0, then the y-intercept cannot be less than 4. Without the restriction that x, y > 0, there is no least value.

3. what about x under restriction x,y>0

4. Then the x-intercept is greater than 1.

5. I can see no point in working out limits on the x and y intercepts separately. The point is the sum of the intercepts.,

Any (non-vertical) line passing through (1, 4) is of the form y= m(x- 1)+ 4 which has y-intercept (set x= 0) equal to 4- m and x-intercept (set y= 0 and solve for x) equal to 1- 4/m. The sum of those intercepts is 4- m+ 1- 4/m= 5- m- 4/m. The derivative, with respect to m, is $-1+ 4/m^2$ which is 0 when $\pm 2$. The second derivative is [/tex]-8/m^3[tex] which is positive (and so gives a minimum) at $m= -2$. With $m= -2$, the sum of intercepts is $4+ 2+ 1+ 2= 9$.

6. sir good evening
sir if we use second value of m=2 we get sum of intercept as 1<9 but double derivative negative ( s should be maximum) where is the problem.

7. When m = 2, the line crosses the x-axis at x = -1. The sum of the intercepts indeed has a local maximum here.

8. If m= 2 the equation of the line is y= 2(x- 1)+ 4. When x= 0, y= -2+ 4= 2 so the y-intercept is 2. When y= 0, 2(x-1)+ 4= 2x+ 2= 0 so x= -1. The sum of the intercepts is 1. But that is a local maximum. If m= -2, the equation of the line is y= -2(x-1)+ 4. When x= 0, y= 2+ 4= 6 so the y-intercept is 6. When y= 0, -2(x- 1)+ 4= -2x+ 6= 0 so x= 3. The sum of the intercepts is 9. But that is a local minimum.

This is because the initial function, 5- m- 4/m, has a discontinuity at x= 0. There is no "connection" between x= -2 and x= 2 so that the local minimum is larger than the local maximum!