1. A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume?

volume= LWH

but 2H+W=3 W=3-2H
and 2H+L=4 or L=4-2H

Put those W, L into the V formula, and take the derivative with respect to H. set to zero, and solve for H

This assumes the box has an open top.

32/2

To find the dimensions of the box with the largest volume, we need to maximize the volume function. Let's break down the problem and solve it step by step:

1. First, visualize the initial sheet of cardboard: a rectangle with dimensions 3 ft. by 4 ft.

2. We need to cut equal-sized squares from each corner of the rectangle. Let's assume the side length of each square is 'x' feet.

3. If we cut squares of side length 'x' from each corner, the resulting dimensions of the rectangle will be (3 - 2x) feet by (4 - 2x) feet.

4. Now, we need to fold up the four edges of the rectangle to form a box. The height of the box will be 'x' feet.

5. Therefore, the dimensions of the box will be (3 - 2x) feet by (4 - 2x) feet by 'x' feet.

6. To find the volume of the box, we multiply these three dimensions: V(x) = (3-2x)(4-2x)x.

7. Expand the equation: V(x) = 4x - 14x^2 + 12x^3.

8. Now, we have a volume function in terms of 'x', and we need to maximize it.

9. To find the maximum, we can take the derivative of the volume function, set it equal to zero, and solve for 'x'.

10. Differentiate V(x) with respect to 'x': V'(x) = 4 - 28x + 36x^2.

11. Set V'(x) = 0 and solve for 'x': 4 - 28x + 36x^2 = 0.

12. Solve the equation using factoring, the quadratic formula, or graphing. In this case, we can use the quadratic formula.

13. Using the quadratic formula, we get two possible values for 'x': x = 0.25 ft and x = 1 ft.

14. We disregard the value x = 0.25 ft because it would result in a negative length for the box, which is not feasible.

15. Therefore, the optimum value for 'x' is x = 1 ft.

16. Substituting x = 1 ft back into the dimensions of the box: (3 - 2x) ft = 3 - 2(1) ft = 3 - 2 ft = 1 ft, and (4 - 2x) ft = 4 - 2(1) ft = 4 - 2 ft = 2 ft.

17. Hence, the dimensions of the box with the largest volume will be 1 ft by 2 ft by 1 ft.

Therefore, by following these steps, we have determined that the box with the largest volume will have dimensions of 1 ft by 2 ft by 1 ft.