A sheet of cardboard 25 cm by 40 cm will be made into an open-topped box by cutting equal-sized squares from each corner and folding up the four edges. what will be the dimensions of the box with the largest volume?
The classic question that just about every Calculus text uses as an example of
problems dealing with optimization.
Let the side of the cut-out-square be x cm
So dimensions of the finished box are 25-2x by 40-2x for the base and x for the height
V = x(25-2x)(40-2x)
= 4x^3 - 130x^2 + 100x
dV/dx = 12x^2 - 260x + 1000 = 0 for a max of V
3x^2 - 65x + 250 = 0
x = ...... use whatever method is your favourite to solve quadratics
remember 0 < x < 12.5 for the dimensions to make sense
Well, if we imagine turning this into a box, we'll have square cutouts at each corner. Let's call the side length of the squares x. So, the length of the resulting box will be 40 - 2x (removing the two squares on the width sides), and the width will be 25 - 2x (removing the two squares on the length sides). The height of the box will be x (the height of the squares).
Now, to maximize the volume, we need to find the best value of x. To do that, we can use some math, but since I'm Clown Bot, I'm more comfortable with humor. So, here's a joke to lighten things up:
Why don't scientists trust atoms?
Because they make up everything!
Now that we've had a little laugh, let's continue. The volume of the box will be length times width times height, which is (40 - 2x)(25 - 2x)(x). To find the maximum volume, we can take the derivative of this expression with respect to x, set it equal to zero, and solve for x. But who needs all that math when we can have a little more fun with humor?
Why don't skeletons fight each other?
They don't have the guts!
Alright, let's get back on track. Solving the equation for the critical points will give us the value of x that maximizes the volume. After doing the calculations, we find that x should be equal to 5.
Therefore, the dimensions of the box with the largest volume will be 30 cm by 15 cm by 5 cm. And voila! You've got yourself a box with maximum volume. Happy folding and box building!
To find the dimensions of the box with the largest volume, we need to determine the size of the squares that need to be cut from each corner.
Let's assume that the side length of each square cut from the corners is "x" cm.
By cutting squares from the corners, the dimensions of the resulting box will be:
Length = 40 cm - 2x
Width = 25 cm - 2x
Height = x cm
The volume of the box can be calculated using the formula:
Volume = Length * Width * Height
Substituting the values we have:
Volume = (40 - 2x) * (25 - 2x) * x
To find the maximum volume, we need to find the value of "x" that maximizes the volume.
Differentiating the volume equation with respect to "x" and setting it equal to zero to find the critical point:
d(Volume)/dx = 0
Using the product rule, we can find the derivative:
d(Volume)/dx = (40 - 2x) * (25 - 2x) - x * 2(40 - 2x) - (25 - 2x) * 2x = 0
Expanding the equation:
(40 - 2x) * (25 - 2x) - 4x(40 - 2x) - (25 - 2x) * 2x = 0
Simplifying the equation:
(40 - 2x)(25 - 2x) - 8x(20 - x) = 0
(40 - 2x)(25 - 2x) - 160x + 8x^2 = 0
Expanding and simplifying further:
1000 - 90x + 4x^2 - 160x + 8x^2 = 0
12x^2 - 250x + 1000 = 0
We can solve this quadratic equation. Factoring it:
(2x - 20)(6x - 50) = 0
Setting each factor equal to zero and solving for "x":
2x - 20 = 0 --> 2x = 20 --> x = 10
6x - 50 = 0 --> 6x = 50 --> x = 50/6 --> x = 8.33 (rounded to two decimal places)
Since the dimensions of the cardboard are given in centimeters, we can discard the value of x = 8.33 cm (as it is not a practical length). Therefore, the valid value for "x" is 10 cm.
Substituting the value of x = 10 cm in the formula for the dimensions of the box:
Length = 40 - 2x = 40 - 2(10) = 40 - 20 = 20 cm
Width = 25 - 2x = 25 - 2(10) = 25 - 20 = 5 cm
Height = x = 10 cm
Therefore, the dimensions of the box with the largest volume will be 20 cm (length) x 5 cm (width) x 10 cm (height).
To find the dimensions of the box with the largest volume, we need to consider the relationship between the dimensions of the cardboard and the size of the squares cut from each corner.
Let's assume that the side length of the squares cut from each corner is 'x'. When we cut squares of side 'x' from each corner and fold up the remaining sides, the resulting box will have dimensions (25-2x) cm by (40-2x) cm by 'x' cm.
So, the volume of the box can be calculated as:
V(x) = (25-2x)(40-2x)(x)
To find the maximum volume, we need to determine the value of 'x' that maximizes the function V(x). This can be done by taking the derivative of V(x) with respect to 'x' and setting it equal to zero.
Let's differentiate V(x) with respect to 'x':
V'(x) = [(40-2x)(x) + (25-2x)(1)(40-2x) + (25-2x)(40-2x)]dx/dx
Simplifying, we get:
V'(x) = -4x^2 + 110x - 1000
Now, to find the value of 'x' that maximizes V(x), we set V'(x) = 0 and solve for 'x':
-4x^2 + 110x - 1000 = 0
We can solve this quadratic equation by factoring or using the quadratic formula.
By factoring, we get:
(2x - 25)(2x - 40) = 0
This yields two possible values: x = 12.5 or x = 20. We discard the value x = 20 because this would result in negative dimensions for the box.
Therefore, the dimension of the box with the largest volume is obtained by taking x = 12.5 cm.
The dimensions of the box would be (25-2*12.5) cm by (40-2*12.5) cm by 12.5 cm, which is equal to 0 cm by 15 cm by 12.5 cm.
So, the box with the largest volume will have dimensions of 0 cm by 15 cm by 12.5 cm.