A rectangular box without a lid is to be made from 12 square meters cardboard. Find the maximum volume of such a box.
The answer is 65 mm because I thought about it for so long and this makes sense
Why did the cardboard go to the gym? To get ripped! Alright, let's calculate the maximum volume of the box. Let's call the length, width, and height of the box L, W, and H, respectively.
Given that the cardboard has an area of 12 square meters, we know that L * W = 12. Since there is no lid, we can assume that the height of the box is equal to the width (H = W).
To find the maximum volume, we need to maximize the product LW * H. Since H = W, we can rewrite this as LW * W.
Now, we need to use a little bit of math magic. Since L * W = 12, we can rewrite the expression as 12/W * W^2.
To find the maximum value, we take the derivative of this expression with respect to W and set it equal to zero. After solving the equation, we find that W equals 2, and L equals 6.
Therefore, the maximum volume of the box is 2 * 6 * 2 = 24 cubic meters.
So, the maximum volume is 24 cubic meters. Now that's a box that can hold a lot of laughter!
To find the maximum volume of a rectangular box without a lid using 12 square meters of cardboard, we need to determine the dimensions of the box that will give us the largest volume.
Let's assume the dimensions of the box are Length (L), Width (W), and Height (H).
Since the box does not have a lid, one side of the cardboard will be missing. Therefore, the area of the box will be given by the sum of the five visible sides: 2(Length x Width) + 2(Length x Height) + Height x Width.
We also know that the area of the cardboard is 12 square meters, so we have the equation:
2(L x W) + 2(L x H) + H x W = 12
To find the maximum volume, we need to maximize the function V = L x W x H.
We can use the method of Lagrange multipliers to find the maximum volume. We define the Lagrangian function as:
L = V - λ(2LW + 2LH + HW - 12)
Where λ is the Lagrange multiplier.
To find the maximum, we take the partial derivatives of the Lagrangian function with respect to L, W, H, and λ and set them to zero:
∂L/∂L = W + 2λ = 0
∂L/∂W = L + 2λ = 0
∂L/∂H = L + λ = 0
∂L/∂λ = 2LW + 2LH + HW - 12 = 0
From the first three equations, we can solve for L, W, and H:
L = -2λ
W = -2λ
H = -λ
Substituting these values into the fourth equation, we have:
2(-2λ)(-2λ) + 2(-2λ)(-λ) + (-λ)(-2λ) - 12 = 0
Simplifying this equation, we get:
8λ^2 + 4λ^2 + 2λ^2 - 12 = 0
14λ^2 = 12
λ^2 = 12/14
λ^2 = 6/7
λ = √(6/7)
Substituting this value back into L, W, and H, we have:
L = -2√(6/7)
W = -2√(6/7)
H = -√(6/7)
Since the dimensions cannot be negative, this solution is not valid.
Therefore, the maximum volume cannot be achieved with a rectangular box without a lid.
To find the maximum volume of the rectangular box, we need to consider the constraints given by the cardboard. We know that the total surface area of the box needs to be 12 square meters.
Let's assume the length, width, and height of the box are L, W, and H, respectively. We need to maximize the volume, V, while keeping the surface area at 12 square meters.
The surface area of the box consists of the following components:
- The top and bottom walls: 2LW
- The side walls: 2LH + 2WH
- The front and back walls: 2WH
From these components, we can write the equation for the total surface area:
2LW + 2LH + 2WH = 12
Now, we need to express the volume V in terms of L, W, and H. The volume of the box is given by:
V = LWH
To maximize the volume, we can use calculus. We need to find the critical points of the volume function V with the constraint equation for the surface area.
First, let's solve the constraint equation for one variable. We can solve it for H:
H = (12 - 2LW) / (2L + 2W)
Now we can substitute this value back into the volume equation:
V = LWH = LW(12 - 2LW) / (2L + 2W)
To find the critical points, we need to find where the derivative of V with respect to L (or W) is equal to zero.
Taking the derivative with respect to L:
dV/dL = (12W - 6LW - 2W^2) / (L + W)^2
Setting this equal to zero and solving for L will give us the critical point(s).
Solving for L, we get:
12W - 6LW - 2W^2 = 0
Rearranging this equation:
L = (12W - 2W^2) / (6W)
L = 2(6 - W)
So, we have found one critical point: L = 2(6 - W).
Now, let's evaluate the volume at the critical point(s) to determine which one(s) give(s) the maximum volume.
V = LWH = (2(6 - W))W(12 - 2(6 - W)) / (2(6 - W) + 2W)
Simplifying the equation, we get:
V = W(6 - W)(12 - 2W) / (6 - W + W)
V = W(6 - W)(12 - 2W) / 6
V = W(6 - W)(12 - 2W) / 6
Now, we can use calculus or algebraic methods to find the maximum value of V. One way to do this is to graph the volume equation and find the highest point on the graph.
By calculating the volume for different values of W, we can find the maximum volume. Remember that the length and width must be positive, so we should confine our search to values of W between 0 and 6.
Alternatively, we can find the maximum value of V by taking the second derivative of the volume equation and evaluating it at the critical point. If the second derivative is negative, we have found a maximum. However, this method may involve more calculations and can be more complex.
By following either of these methods, you can find the maximum volume of the rectangular box.
This problem involves three variables, length l, width w, and height h.
They can be reduced to two using the constraint that the surface area is 12 sq. m.
Out of the three, w and l are symmetrical, so assumption can be made that when w=l the volume is either a maximum or a minimum.
Assume therefore w=l and proceed with the volume calculation:
V=wlh
subject to 2h(w+l)+wl=12
When l=w, this reduces to
2h(2w)+w²=12
from which
h=(12-w²)/4w
Substitute in V:
V=wlh
=w²h
=w(12-w²)/4
=(12w-w³)/4
At maximum (or minimum) volume,
dV/dw=3-3w²/4=0
w²=4
w=2 (metres)
Check maximum or minimum:
d²V/dw²=-6w/4 <0 => maximum.
Therefore the dimensions of the box should be:
w=l=2m h=12/2²=3m.
If the assumption of symmetry cannot be made, Lagrange multipliers can be used to determine the three dimensions.