If an open box is made from a tin sheet 7 in. square by cutting out identical squares from each corner and bending up the resulting flaps, determine the dimensions of the largest box that can be made. (Round your answers to two decimal places.)
Height:
Length:
Width:
base = 7- 2x
height = x
volume = x(7-2x)^2
= 49x - 28x^2 + 4x^3
d(volume)/dx = 49 - 56x + 12x^2
= 0 for a max volume
12x^2 - 56x + 49 = 0
x = (56 ± √784)/24
= (56 ± 28)/24 = 3.5 or 7/6 or 1.1666...
but clearly x < 3.5 or we have cut the whole base away.
base is 7 - 2(7/6) = 14/3 by 14/3
and the height is 7/6
round to your required decimals
new length = 7 - 2 h
height = h
volume = (7-2h)(7-2h)(h)
v = (49 -28 h + 4 h^2)h
so
v = 4 h^3 -28 h^2 + 49 h
dv/dh = 0 for max or min
dv/dh = 12 h^2 -56 h + 49 = 0
(6h -7)(2 h -7) = 0
h = 7/6 or h = 7/2
if h = 7/2, the box has zero bottom
so the answer is h = 7/6
7 - 2(7/6) = 7-7/3 = 14/3 = length and width
To determine the dimensions of the largest box that can be made, we need to find the value of the side length that maximizes the volume of the box.
Let's denote the side length of the squares cut out from each corner as "x". This means that the length of the resulting flaps would also be "x". Therefore, the height of the box would also be "x", and the dimensions of the base of the box would be (7-2x) by (7-2x).
The volume V of the box can be calculated by multiplying the base area by the height:
V = (7-2x) * (7-2x) * x
V = x(49 -14x + 4x^2)
V = 4x^3 - 14x^2 + 49x
To maximize the volume, we need to find the value of x that makes the derivative of V equal to 0. So, let's differentiate V with respect to x:
dV/dx = 12x^2 - 28x + 49
Setting dV/dx equal to 0 and solving for x:
12x^2 - 28x + 49 = 0
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
x = (-(-28) ± √((-28)^2 - 4(12)(49))) / (2(12))
x = (28 ± √(784 - 2352)) / 24
x = (28 ± √(-1568)) / 24
Since the value inside the square root is negative, there are no real solutions for x. This indicates that the volume of the box is maximized when x is at one of the endpoints of its domain, which is [0, 7/2].
We evaluate the volume of the box at these endpoints:
V(0) = 4(0)^3 - 14(0)^2 + 49(0) = 0
V(7/2) = 4(7/2)^3 - 14(7/2)^2 + 49(7/2) = 49/4
From the evaluations, we can see that the volume is maximized when x = 7/2.
Therefore, the dimensions of the largest box that can be made are:
Height: 7/2 inches
Length: 7 - 2(7/2) = 7 - 7 = 0 inches
Width: 7 - 2(7/2) = 7 - 7 = 0 inches
The width and length of the box would both be zero, which indicates that the largest box would just be a flat piece of tin sheet with flaps folded upwards to form the sides.
To determine the dimensions of the largest box that can be made, we need to find the dimensions that will maximize the volume of the box.
Let's start by visualizing the situation. We have a square tin sheet with a side length of 7 inches. We will cut identical squares from each corner and fold the resulting flaps to create the box.
Let's assume that we cut squares of length x from each corner. This means that the side length of the resulting box will be (7 - 2x) inches, as we are removing x inches from each side.
To find the height of the box, we need to determine the length of the flap that will be folded up. This length will also be x inches.
With these dimensions in mind, we can calculate the volume of the box. The formula for volume is V = length x width x height.
Volume = (7 - 2x) * (7 - 2x) * x
To find the maximum volume, we need to maximize this function. We can do this by finding the critical points of the function, which occur when the derivative is equal to zero.
Let's differentiate the volume function with respect to x:
dV/dx = 4(7 - 2x) * x + (7 - 2x)^2
Simplifying this expression:
dV/dx = 4x(7 - 2x) + (7 - 2x)(7 - 2x)
Next, set this derivative equal to zero:
4x(7 - 2x) + (7 - 2x)(7 - 2x) = 0
Solve this equation to find the critical points.
Once you find the critical points, you can plug them back into the volume function to determine the maximum volume. The dimensions of the resulting box will correspond to these critical points.
By solving this equation, you should be able to find the values of x that maximize the volume. The maximum volume will correspond to the dimensions of the largest box that can be made from the tin sheet.
Height: So the corners are cut out from a square sheet, which means the height will be the same as the length of each cut-out square. Let's call the length of each cut-out square x. When the flaps are bent up, the height is x as well.
Length: After the corners are cut out, the remaining sheet will have a length of 7 - 2x (two corners are cut out from each end). When folded, this length will be the same as the length of the box.
Width: Similarly, the width of the box will be the same as the width of the remaining sheet, which is also 7 - 2x.
To maximize the volume, we need to maximize the length, width, and height of the box. So let's differentiate their product with respect to x and find the value of x that yields the maximum volume.
V = (7 - 2x)(7 - 2x)x
Differentiating with respect to x:
dV/dx = 2(7 - 2x)(-2x) + (7 - 2x)(7 - 2x)
Setting dV/dx = 0:
2(7 - 2x)(-2x) + (7 - 2x)(7 - 2x) = 0
Expanding and simplifying:
-4x(7 - 2x) + (7 - 2x)(7 - 2x) = 0
-28x + 8x^2 + 49 - 28x + 4x^2 = 0
12x^2 - 56x + 49 = 0
Solving this quadratic equation, we get two values for x:
x = 1.71 and x = 2.88
Since the dimensions cannot be negative, we discard x = 2.88.
Therefore, the largest box that can be made has the following dimensions:
Height: 1.71 inches
Length: (7 - 2(1.71)) = 3.58 inches
Width: (7 - 2(1.71)) = 3.58 inches
Now that we have the dimensions, we can enjoy a perfectly sized box for all our tin-related storage needs. Just try not to cut yourself on those flaps!