# Minimum value problem

• Jan 19th 2009, 11:32 AM
gebbal
Minimum value problem
If y=2x-8, what is the minimum value of the product xy?

i only know how to get the max, i know you need to find the dirivative but i dont know how to make it the minimum.
• Jan 19th 2009, 12:24 PM
Jester
Quote:

Originally Posted by gebbal
If y=2x-8, what is the minimum value of the product xy?

i only know how to get the max, i know you need to find the dirivative but i dont know how to make it the minimum.

So you have

\$\displaystyle A = x (2x - 8)\$

A will have a max. (i.e. the derivative = 0). For a minimum, you need to check the endpoints. I used A as it might be an area and so \$\displaystyle [0,4]\$ and check \$\displaystyle x = 0,\;\;\;x=4\$.
• Jan 19th 2009, 12:54 PM
gebbal
im still confused. is the answer 0? since thats one of the end points.
• Jan 19th 2009, 03:02 PM
Jester
Quote:

Originally Posted by gebbal
im still confused. is the answer 0? since thats one of the end points.

Let us ask, where did this problem arise?
• Jan 19th 2009, 03:55 PM
HallsofIvy
The problem is to minimize \$\displaystyle A= 2(x^2- 4x)\$. You don't need to differentiate at all. Completing the square, [tex]A= 2(x^2- 4x+ 4- 4)= 2(x^2- 4x+ 4)- 8= 2(x-2)^2- 8[/itex]. That is a parabola opening upwards so its minimum value is at the vertex, (2, -8).

If you must differentiate, A'= 4x- 8= 0 so x= 2. gebbal, you seem to be under the impression that the derivative is 0 only at a maximum. A derivative is 0 at a maximum, minimum or "saddle" point (the derivative of \$\displaystyle x^3\$, \$\displaystyle 3x^2\$ is 0 at x= 0 which is neither a maximum nor a minimum). After finding where the derivative is 0 (a "critical point"), you still need to determine whether a maximum or minimum or saddle point. You can do that by checking the first derivative around the point. Here, A'= 4(x-2). If x< 2, that is negative and if x> 2, that is positive. That is, A is decreasing for x< 2 and then increases: x= 2 gives a minimum.

Or you can check the second derivative: A"= 4> 0. Since A" is positive, A' is increasing: it goes from negative to positive so, as before, x= 2 gives a minimum. Oh, and that minimum value is, as we saw by completing the square, A(2)= 2(2(2)- 8)= 2(-4)= -8.