a manufacturer of open tin boxes wishes to make use of tin with dimension 10 inches by 20 inches by cutting equal squares from the four corners and turning up sides. find the lengths of the side of the square to be cut out if an open box having the largest possible volume is to be obtained from each piece of the tin.

v = x(10-2x)(20-2x)

dv/dx = 4(3x^2-30x+50)
Now just find where dv/dx=0 and decide whether that's a minimum or a maximum.

To find the length of the side of the square to be cut out, we need to maximize the volume of the open box formed after cutting.

Let's assume that the side length of the square to be cut out is "x" inches.

When the square is cut out from each corner of the tin, the dimensions of the resulting box will be:
Length = 20 - 2x inches
Width = 10 - 2x inches
Height = x inches (since the square is turned up to form the open box)

The volume of the box can be calculated by multiplying these dimensions together:
Volume = Length * Width * Height
= (20 - 2x) * (10 - 2x) * x

To maximize the volume, we can take the derivative of the volume expression with respect to x and set it equal to zero. Then, solve for x.

First, let's expand the volume expression:
Volume = (20 - 2x)(10 - 2x)x
= x(20 - 2x)(10 - 2x)

Taking the derivative of the volume expression with respect to x:
dV/dx = (20 - 2x)(10 - 2x) + x(-2)(10 - 2x) + x(20 - 2x)(-2)
= (20 - 2x)(10 - 2x) - 2x(10 - 2x) - 2x(20 - 2x)

Setting dV/dx equal to zero:
(20 - 2x)(10 - 2x) - 2x(10 - 2x) - 2x(20 - 2x) = 0

Now, solve this equation to find the length of the side of the square to be cut out.