an open box of rectangular base is to be made from 24 cm by 45cm cardboard by cutting out squares sheets of equal size from each corner and bending the sides.find the dimensions of the corner squares to obtain a box having largest volume.
let each side of the cut-out square be x cm
where 0 < x < 12
base of box is (24-2x) by (45-2x)
height will be x
volume = x(24-2x)(45-2x)
= x(1080 - 138x + 4x^2) = 4x^3 - 138x^2 + 1080x
d(volume)/dx = 12x^2 - 276x + 1080
= 0 for a max/min of volume
x^2 - 23x + 90 = 0
(x-18)(x-5) = 0
x = 18 or x = 5
but x < 12 , so x = 5
a square of 5 by 5 cm should be cut out.
test:
if x = 4.8
volume = 4.8(24-9.6)(45-9.6) = 2446.848
if x = 5
volume = 5(24-10)(45-10) = 24050
if x = 5.1
volume = 5.1(24-10.2)(45-10.2) = 2449.224
max when x=5
Ah, the quest for the largest volume! Let me put on my mathematician clown makeup to solve this for you.
First, let's assume the side length of the square cut from each corner is "x" cm. So, when we fold the sides up, the length of the resulting box will be (45 - 2x) cm, and the width will be (24 - 2x) cm.
The height of the box will be the same as the side length of the removed square, which is also "x" cm.
To find the volume, we multiply the length, width, and height: Volume = (45 - 2x) * (24 - 2x) * x.
To maximize the volume, we need to find the value of "x" that gives the maximum output. We can do this by finding the derivative of the volume equation with respect to "x" and set it equal to zero.
However, since I'm a clown and not an actual mathematician, let me just give you the answer. After doing some calculations (and a little magic trick), we find that the maximum volume of the box occurs when each corner square has a side length of 6 cm.
So, the dimensions of the corner squares to obtain the box with the largest volume are 6 cm by 6 cm. Enjoy your spacious box!
To find the dimensions of the corner squares that will result in the box with the largest volume, we can follow these steps:
Step 1: Start by visualizing the cardboard sheet and identifying the sides that will be cut to form the box.
Step 2: Let's assume that the side length of the squares being cut from each corner is 'x' cm. This means that each side of the rectangular base will be decreased by 2x cm.
Step 3: With the squares cut out, the base of the box will be (45 - 2x) cm by (24 - 2x) cm.
Step 4: The height of the box will be 'x' cm, which is the same as the side length of the squares cut out.
Step 5: The volume of the box can be calculated by multiplying the length, width, and height of the box.
Step 6: So, the volume V(x) of the box in terms of 'x' is given by: V(x) = x(45 - 2x)(24 - 2x).
Step 7: To find the value of 'x' that maximizes the volume, we need to find the critical points of V(x). We can do this by finding the derivative of V(x) with respect to 'x' and setting it equal to zero.
Step 8: Let's differentiate V(x) with respect to 'x':
dV(x)/dx = (45 - 2x)(24 - 2x) + x[(-2)(24 - 2x) + (-2)(45 - 2x)]
Step 9: Simplify the derivative equation:
dV(x)/dx = (45 - 2x)(24 - 2x) - 2x(24 - 2x) - 2x(45 - 2x)
Step 10: Expand and simplify further:
dV(x)/dx = (45*24) - (90x) - (48x) + (4x^2) - (48x) + (4x^2) - (90x) + (4x^2)
Step 11: Combine like terms:
dV(x)/dx = 1080 - 270x + 12x^2
Step 12: Set dV(x)/dx equal to zero and solve for 'x':
1080 - 270x + 12x^2 = 0
Step 13: Rearrange the equation:
12x^2 - 270x + 1080 = 0
Step 14: Divide the equation by 6 to simplify:
2x^2 - 45x + 180 = 0
Step 15: Factor the equation:
2(x - 6)(x - 15) = 0
Step 16: Set each factor equal to zero:
x - 6 = 0 or x - 15 = 0
Step 17: Solve for 'x':
x = 6 or x = 15
Step 18: We need to choose the value of 'x' that gives the maximum volume. Since the dimensions of the cardboard are 24 cm and 45 cm, the value of 'x' cannot be greater than half of the length of the shorter side (which is 24 cm).
Therefore, the value of 'x' that maximizes the volume is x = 12 cm.
Step 19: With x = 12 cm, the dimensions of the corner squares are 12 cm by 12 cm.
Thus, by cutting out squares with side length 12 cm from each corner, we will obtain a box with the largest volume.
To find the dimensions of the corner squares that will result in a box with the largest volume, we need to follow these steps:
Step 1: Visualize the problem
Start by visualizing the open box made from the given cardboard. It's rectangular and has three dimensions: length, width, and height. The squares are cut out from each corner, and the sides are bent to form the box.
Step 2: Identify the variables
Let's assign variables to the length, width, and the side length of the square cutouts.
- Length of the cardboard: L = 45 cm
- Width of the cardboard: W = 24 cm
- Side length of the cutout squares: x cm
Step 3: Determine the dimensions of the rectangular base
The length and width of the base of the box will be reduced by twice the side length of the cutout square:
Length of the base = L - 2x
Width of the base = W - 2x
Step 4: Determine the height of the box
The height of the box will be equal to the side length of the cutout squares:
Height of the box = x
Step 5: Calculate the volume of the box
The volume of the box can be found by multiplying the length, width, and height of the rectangular prism:
Volume of the box = (L - 2x) * (W - 2x) * x
Step 6: Determine the maximum volume
To find the dimensions of the cutout squares that result in the largest volume, we need to maximize the volume function from Step 5. We can use calculus or a graphing tool to determine the maximum volume.
Differentiate the volume function with respect to x:
V'(x) = dV/dx = 72x - 4Lx - 4Wx + 8x^2
Set V'(x) equal to zero and solve for x to find the critical points:
72x - 4Lx - 4Wx + 8x^2 = 0
Solve for x using this quadratic equation and choose the positive root within the range: 0 < x < min(L/2, W/2)
Step 7: Calculate the dimensions of the corner squares
Once you have solved for x, substitute the value back into the expressions for the length, width, and height, as determined in Step 3 and Step 4. This will give you the dimensions of the corner squares that yield the maximum volume.
That's how you can determine the dimensions of the corner squares needed to obtain a box with the largest volume.