A = 2x^2 + 20/x

"Hence determine the value of x that will give the minimum surface area to be coated."

I don't even know what kind of graph I'm working with here, but I guess I'm looking for a local minimum? Any hints? Thanks

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- August 5th 2012, 05:32 AMPulsysFinding minimum value
A = 2x^2 + 20/x

"Hence determine the value of x that will give the minimum surface area to be coated."

I don't even know what kind of graph I'm working with here, but I guess I'm looking for a local minimum? Any hints? Thanks - August 5th 2012, 06:08 AMProve ItRe: Finding minimum value
- August 5th 2012, 06:16 AMPulsysRe: Finding minimum value
- August 5th 2012, 06:19 AMProve ItRe: Finding minimum value
You should know that functions will either be maximised at turning points or endpoints. The function is discontinuous at x = 0, so it should be clear that when you approach 0 from the left, the function approaches , so the function does not have a minimum.

- August 5th 2012, 06:46 AMPulsysRe: Finding minimum value
x refers to the dimensions of an object, so the domain would be x>0. So I'm looking for the stationary point to the right of the y-axis (am I wrong to call that a "local minimum"?). How could I do this? I assume I could use the derivative of the function to find the point where the gradient is 0. Assuming I've got the right idea, could anyone be so kind as to give me the derivative of this function?

- August 5th 2012, 06:49 AMProve ItRe: Finding minimum value
- August 5th 2012, 07:07 AMPulsysRe: Finding minimum value
Haha yes, that's my bad, very sorry about the wasted time. I omitted the rest of the question to avoid wasting everyone's time as I originally thought it'd be irrelevant. Here's the former part of the question:

A manufacturer wants to dip an open box in plastic resin to coat the inside and the outside of the box, including the base, with plastic resin. The box has a square base. The sides of the square base measure x metres each. - August 5th 2012, 07:28 AMProve ItRe: Finding minimum value
Well, you are correct that you would have to use the derivative. To find the derivative, it would help if you wrote the function as ...