A 5 cm by 5 cm square is cut from each corner of a rectangular piece of cardboard. The sides are folded up to make an open box with a maximum volume. If the perimeter of the base is 50 cm, what are the dimensions of the box?
If the cardboard had dimensions x and y, then
2(x-10 + y-10) = 50
x+y = 45
Now, the volume is
v = 5(x-10)(y-10) = 5(x-10)(35-x) = -5(x^2-45x+350)
The vertex of this parabola (maximum volume) is at (45/2 , 3125/4)
To solve this problem, we need to find the dimensions of the rectangular piece of cardboard that will result in the maximum volume of the open box.
Let's assume the length of the rectangular piece of cardboard is "L" and the width is "W".
When we cut a 5 cm square from each corner, the remaining length and width of the base of the box will be (L - 10) cm and (W - 10) cm respectively because 5 cm is cut from both ends.
The height of the box will be 5 cm because the squares cut from the corners determine the height.
Now, we need to find the dimensions of the rectangular piece of cardboard that will maximize the volume.
The volume of a rectangular box is given by V = Length × Width × Height.
So, the volume of the box in terms of the given dimensions L and W will be:
V = (L - 10) × (W - 10) × 5
Next, we need to find the value of L and W that satisfies the given conditions - the perimeter of the base is 50 cm.
The perimeter of a rectangle is given by P = 2(L + W).
If the perimeter of the base is 50 cm, then:
2(L - 10) + 2(W - 10) = 50
Simplifying the equation:
2L - 20 + 2W - 20 = 50
2L + 2W - 40 = 50
2L + 2W = 90
Now, we have two equations:
V = (L - 10) × (W - 10) × 5
2L + 2W = 90
To find the dimensions of the box, we need to find the values of L and W that maximize the volume using these equations.
One way to solve this is by substituting one equation into the other.
From the perimeter equation, we have:
L = (90 - 2W) / 2
Substituting this value of L into the volume equation, we get:
V = ((90 - 2W) / 2 - 10) × (W - 10) × 5
Now, we have the volume equation in terms of W only.
To maximize the volume, we can take the derivative of the volume equation with respect to W, set it equal to zero, and solve for W.
Once we find the value of W, we can substitute it back into the equation to find the value of L.
Finally, we can plug in the values of L and W into the perimeter equation to make sure the perimeter of the base is 50 cm.
However, due to the complexity of the calculations involved, it would be easier to use a graphing calculator or software to plot the volume equation and find the maximum point on the graph, which will give us the values of L and W that maximize the volume.
With these values, we can determine the dimensions of the box.
To find the dimensions of the box, we can start by finding the length of the base of the box.
Let's assume the length of the base is x cm.
Since the perimeter of the base is 50 cm, the sum of all the sides of the base is 50 cm.
So, 2 times the length plus 2 times the width of the base is equal to 50 cm.
Therefore,
2x + 2x = 50 cm
4x = 50 cm
x = 12.5 cm
Now, to find the dimensions of the box, we need to consider that each side of the base will be decreased by 5 cm (2 times 5 cm) due to the square cut from each corner.
So, the length of the box will be (12.5 cm - 5 cm - 5 cm) = 2.5 cm.
Similarly, the width of the box will also be (12.5 cm - 5 cm - 5 cm) = 2.5 cm.
Therefore, the dimensions of the box will be 2.5 cm by 2.5 cm.