# intercept

• Jun 11th 2011, 08:11 AM
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)
• Jun 11th 2011, 08:38 AM
emakarov
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.
• Jun 11th 2011, 08:50 AM
what about x under restriction x,y>0
• Jun 11th 2011, 08:57 AM
emakarov
Then the x-intercept is greater than 1.
• Jun 11th 2011, 09:52 AM
HallsofIvy
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 $4+ 2+ 1+ 2= 9$.
• Jun 11th 2011, 10:28 AM