All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by s(x)equalsx squared plus startfraction 240 over x endfraction , where x is the length of the sides of the base. find the absolute minimum of the surface area function on the interval (0,infinity). what are the dimensions of the box with minimum surface area?

The absolute minimum of the surface area function on the interval is

The dimensions of the box with minimum surface area are: the base edge and the height

Step-by-step explanation:

We are given the surface area of a box where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area () and find the critical points of the derivative ().

Next,

There is a undefined solution and a real solution . These point divide the number line into two intervals and

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before , increases after it, and is defined at . So f(x) has a relative minimum point at .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for .

and for we get:

Therefore S(x) has a minimum at which is:

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 and the volume of a box with a square base is , we solve for h

Substituting V = 60 and

The dimension are the base edge and the height

what? ?

what type of question is thiss

step-by-step explanation:

3; 9 is 8 to -4

step-by-step explanation: