Math Help - Minimum value problem

1. 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.

2. 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

$A = x (2x - 8)$

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 $[0,4]$ and check $x = 0,\;\;\;x=4$.

3. im still confused. is the answer 0? since thats one of the end points.

4. 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?

5. The problem is to minimize $A= 2(x^2- 4x)$. 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 $x^3$, $3x^2$ 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.