You were close, but the question asks for price minimisation, not surface area minimisation.
Make the substitution:
Set P'(x) to 0 and solve for x.
This question really threw me for a loop:
A closed rectangular container with a square base is to have a volume of 2250 cubic inches. The material for the top and bottom of the container will cost $2 per square inch and the material for the sides will cost $3 per square inch. Find the dimensions of the container of least cost.
What I have so far:
But when I thought about trying to find the feasible domain, I got stumped. The question doesn't specify a minimum or maximum value for either a side of the base or the height of the container. So, theoretically, x would get infinitely large as h got infinitely small and vice versa. Is there a generally accepted thing to do in this situation or am I doing something wrong here?
Thanks for the help
Oh, and i had a few more equations relative to the problem:
I separated the equation for surface area into where S is the total surface area, is the area of the base( ), and is the area of a side( ).
I then made the equation where P is the total price. But I didn't know what to do with all this...
Edit: I suppose I'll just attempt to finish the problem with a feasible domain of 0 < x < , although I won't be able to test for the endpoints... oh well.