# An open -top box is to be made by cutting small congruent squares from the corners of a 12-by12-in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible?

By signing up, you agree to our Terms of Service and Privacy Policy

To maximize the volume of the box, we need to maximize the area of the base while also maximizing the height.

Let (x) represent the length of each side of the squares cut from the corners. When these squares are removed and the sides are folded up, the dimensions of the resulting box will be (12-2x) by (12-2x) by (x).

The volume of the box can be expressed as (V(x) = x(12-2x)^2).

To find the value of (x) that maximizes (V(x)), we take the derivative of (V(x)) with respect to (x), set it equal to zero, and solve for (x).

(V'(x) = 12(12-2x)^2 - 2x(12-2x)(-2) = 0)

After simplifying and solving, we find (x = 3).

Thus, the squares cut from the corners should be (3) inches by (3) inches each to maximize the volume of the box.

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the linearization of #F(x) = cos(x)# at a=pi/4?
- The corners are removed from a sheet of paper that is #3# ft square. The sides are folded up to form an open square box. What is the maximum volume of the box?
- What is the local linearization of #e^sin(x)# near x=1?
- An arched window (an upper semi-circle and lower rectangle) has a total perimeter of #10 \ m#. What is the maximum area of the window?
- What is the rate of change of the width (in ft/sec) when the height is 10 feet, if the height is decreasing at that moment at the rate of 1 ft/sec.A rectangle has both a changing height and a changing width, but the height and width change so that the area of the rectangle is always 60 square feet?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7