open top rectangular box made from 35 x 35 inch piece of sheet metal by cutting out equal size squares from the corners and folding up the sides. what size squares should be removed to produce box with maximum volume.
volume= l*w*h
2h+w=35 or w=35-2h
2h+l=35 or l= 35-2h
volume= (35-2h)(35-2h)h
dV/dh= 2(35-2h)(-2)h+ (35-2h)^2 =0
4h=35-2h
6h=35
solve for h, then the cut squares are hxh.
hen the corners of size x are cut out, the dimensions of the box are
35-2x and 35-2x and x, and the volume is thus
v = (35-2x)(35-2x)x
= 1225x - 140x^2 + 4x^3
dv/dx = 1225 -280x + 12x^2
dv/dx = 0 when x = 5.83333
an open box is to be made from a piece of metal 16 by 30 inches by cutting out squares of equal size from the corners and bending up the sides. what size should be cut out to create a box with the greatest volume? what is the maximum volume?
To find the size of squares that should be removed to produce a box with the maximum volume, let's break down the problem step by step.
1. Visualize the box: Start by considering the original sheet of metal with dimensions of 35 inches by 35 inches. Cut squares of equal size from each corner and fold up the sides to form an open-top rectangular box.
2. Define the variables: Let's assume that the size of the square cut out from each corner has an edge length of 'x' inches.
3. Determine the dimensions of the resulting box: After cutting and folding, the original 35-inch sides will decrease by 2x inches (x on each side) due to the corners being removed. This means the new dimensions of the bottom of the box will be (35 - 2x) inches by (35 - 2x) inches. The height of the box will be 'x' inches.
4. Calculate the volume of the box: The volume of a rectangular box is given by the formula V = length × width × height. In this case, we have V = (35 - 2x) × (35 - 2x) × x = 4x^3 - 140x^2 + 1225x.
5. Find the critical points: To determine the value of 'x' that maximizes the volume, we need to find the critical points of the volume function. Take the derivative of the volume function with respect to 'x' and set it equal to zero:
dV/dx = 12x^2 - 280x + 1225 = 0.
6. Solve for 'x': Solve the quadratic equation 12x^2 - 280x + 1225 = 0 to find the values of 'x' that give us critical points.
7. Determine the optimal 'x' value: From the solutions obtained in step 6, select the positive value of 'x' within the feasible range (0 < x < 17.5). This value will correspond to the size of the square that should be removed to achieve the maximum volume.
8. Calculate the maximum volume: Substitute the optimal 'x' value into the volume equation V = 4x^3 - 140x^2 + 1225x to find the maximum volume the box can have.
By following these steps, you will find the size of squares that should be removed to produce a box with the maximum volume.