# calculus optimizationA supermarket employee wants to construct an open-top box from a 16 by 30 in. piece of cardboard. To do this, the employee plans to cut out squares of equal size from the four...

calculus optimization

A supermarket employee wants to construct an open-top box from a 16 by 30 in. piece of cardboard. To do this, the employee plans to cut out squares of equal size from the four corners so the four sides can be bent upwards. What size should the squares be in order to create a box with the largest possible volume?

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Let x be the size of the side of the squares to be cut out of the corners (x measured in inches.)

Once folded, the parallelpiped will have dimensions x, 16-2x, and 30-2x.

Thus the volume V is:

`V=x(16-2x)(30-2x)=4x^3-92x^2+480x`

In order to find extrema, we find the first derivative. The critical points are located where the first derivative is zero or fails to exist. We then check eachof the critical points to determine if they are relative maximums, minimums, or neither.

`(dV)/(dx)=12x^2-184x+480` . Setting the derivative equal to zero yields:

`4(3x-10)(x-12)=0==>x=10/3,x=12`

x=12 cannot be a solution in the physical situation since one of the sides is only 16 inches long. `x=10/3` is a local maximum (using the first derivative test).

**The size of the square to cut is a square with side length `10/3 "in"` .**