A box without a top is to be made from a rectangular piece of cardboard, with dimensions 9 in. by 10 in., by cutting out square corners with side length x and folding up the sides.
What is the optimal volume of the box, and what is the optimal cut?

The optimal volume is 63.1142 in3, with a cut of 1.5767 inches

Step-by-step explanation:

If we make a cut of x inches to create the box, the dimensions of the box will be:

Length = 10 - 2x

Width = 9 - 2x

Height = x

So the volume of the box would be:

Volume = (10 - 2x) * (9 - 2x) * x

Volume = 4x3 - 38x2 + 90x

To find the maximum volume, we need to take the derivative of the volume and find the values of x where it is equal to zero:

dV/dx = 12x2 - 76x + 90 = 0

6x2 - 38x + 45 = 0

Using Bhaskara's formula, we have:

Delta = 38^2 - 4*6*45 = 364

sqrt(Delta) = 19.08

x1 = (38 + 19.08) / 12 = 4.7567

x2 = (38 - 19.08) / 12 = 1.5767

Testing these values in the volume equation, we have:

Volume1 = 4*(4.7567)^3 - 38*(4.7567)^2 + 90*(4.7567) = -1.1883

(Negative value for the volume is not valid)

Volume1 = 4*(1.5767)^3 - 38*(1.5767)^2 + 90*(1.5767) = 63.1142

So the optimal volume is 63.1142 in3, with a cut of 1.5767 inches

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation: