1. A long strip of copper 8 inches wide is to be made into a rain gutter by turning up the sides to form a trough with a rectangular cross section. Find the dimentions of the cross-section if the carrying capacity of the trough is to be a maximum.
2. A sheet of cardboard 18 in square is used to make an open box by cutting squares of equal size from the corners and folding up the sides. What size squares should be cut to obtain a box with largest possible volume?
Please help! Thank you!
A long strip of copper 8 inches wide is to be made into a rain gutter by turning up the sides to form a trough with a rectangular cross section. Find the dimentions of the cross-section if the carrying capacity of the trough is to be a maximum.
If x = the width, then the height y = (8 - x)/2
The volume of the crossection is therefore V = x(8 - x)/2 = 4x - x^2/2
Setting the first derivitive dv/dx = 4 - x = 0, x = 4 making y = 2
2. A sheet of cardboard 18 in square is used to make an open box by cutting squares of equal size from the corners and folding up the sides. What size squares should be cut to obtain a box with largest possible volume?
If the squares are x by x, the volume becomes V - (18 - 2x)(18 - 2x)
Can you take it from here?
if a boy gives his sister 2 of his years to make her twice his age, then he gives her 3 years to make her three times his age, how old are both of the kids
1. Ah, the quest for the perfect rain gutter! To maximize its carrying capacity, we'll need to optimize the dimensions of the cross-section. Let's call the length of the base of the rectangular cross-section "x." Now, given that the strip of copper is 8 inches wide, we can conclude that the width of the base must be 8 - 2x (since we're turning up the sides).
To find the carrying capacity, we need to maximize the area of the cross-section. The area, A, is equal to the length times the width. Therefore, A = x * (8 - 2x).
To maximize A, we find the critical points by taking the derivative of A with respect to x. That gives us dA/dx = 8 - 4x. Setting this equal to zero, we find that x = 2.
So, the dimensions of the cross-section that will maximize the carrying capacity are: length = 2, width = 8 - 2(2) = 4 inches. Voila! You're now a master gutterologist! 🌧️
2. Ah, the art of cardboard origami! To create the largest possible volume for your open box, we need to be precise with our folding strategy. Let's assume that the length of the squares we cut from the corners is "x."
When we fold up the sides of the cardboard, the height of the box will be equal to the length of the squares that were cut off, "x." The width will be equal to the original length of the cardboard minus twice the length of the squares, 18 - 2x. The length will be equal to the original width minus twice the length of the squares, also 18 - 2x.
To calculate the volume, V, we just multiply the length, width, and height together: V = x(18 - 2x)(18 - 2x).
To find the maximum volume, we'll again take the derivative of V with respect to x. This is going to be a bit trickier than folding the cardboard! Taking the derivative, we get dV/dx = 36x - 72 - 4x^2.
Setting this equal to zero, we find that x = 3. If you cut squares with sides of 3 inches, you'll achieve the largest possible volume for your box.
And there you have it! You're now a cardboard connoisseur, creating boxes that are the talk of the town. 📦
1. To find the dimensions of the cross-section of the trough that will maximize its carrying capacity, we need to maximize the volume of the rectangular cross-section.
Let's assume the width of the rectangular cross-section is x inches. The height of the cross-section can be denoted as h inches.
Since the copper strip is 8 inches wide, we know that the length of the rectangular cross-section will be 8 - 2x (2x because we need to fold up the sides).
The volume V of the rectangular cross-section can be calculated using the formula:
V = length * width * height
V = (8 - 2x) * x * h
To find the maximum carrying capacity, we need to find the maximum volume of the cross-section. We can do this by taking the derivative of the volume equation with respect to x and finding the critical points.
Let's differentiate the volume equation with respect to x:
dV/dx = (8 - 4x)h
Setting dV/dx = 0, we get:
8 - 4x = 0
4x = 8
x = 2
Now that we have the value of x, we can substitute it into the volume equation to find the value of h.
V = (8 - 2x) * x * h
V = (8 - 2*2) * 2 * h
V = 4h
Therefore, the maximum carrying capacity of the trough is achieved when the cross-section dimensions are 2 inches by 4h inches.
2. To find the size of the squares that should be cut from the corners of a square cardboard sheet to obtain a box with the largest possible volume, we need to maximize the volume of the box.
Let's assume we cut squares of size x inches from each corner of the cardboard sheet. The resulting length of the sides of the box will be 18 - 2x inches.
The height of the box will be x inches since we fold up the sides.
The volume V of the box can be calculated using the formula:
V = length * width * height
V = (18 - 2x) * (18 - 2x) * x
To find the maximum volume, we need to take the derivative of the volume equation with respect to x and find the critical points.
Let's differentiate the volume equation with respect to x:
dV/dx = 4x^2 - 72x + 324
Setting dV/dx = 0, we get:
4x^2 - 72x + 324 = 0
To solve this quadratic equation, we can use factoring or the quadratic formula.
Using factoring:
4(x-9)(x-9) = 0
This gives us two solutions:
x-9 = 0 or x-9 = 0
x = 9 or x = 9
Since x represents the size of the squares that need to be cut, we can only have positive values for x. Therefore, the only valid solution is x = 9.
Substituting x = 9 into the volume equation, we can find the maximum volume:
V = (18 - 2x) * (18 - 2x) * x
V = (18 - 2*9) * (18 - 2*9) * 9
V = 6 * 6 * 9
V = 324
Therefore, the size of the squares that should be cut from the corners of the cardboard sheet to obtain a box with the largest possible volume is 9 inches. The resulting box will have a volume of 324 cubic inches.
1. To find the dimensions of the cross-section that will maximize the carrying capacity of the trough, we can use the concept of optimization.
Let's start by visualizing the cross-section of the trough. Since the strip of copper is 8 inches wide, we can assume that the height of the cross-section is h inches. The width of the base of the cross-section will then be 8 - 2h inches (subtracting h from each side to account for the upturned sides).
The carrying capacity of the trough is determined by the cross-sectional area. In this case, it is the product of the width and height of the cross-section, which is (8 - 2h) * h = 8h - 2h^2 square inches.
To find the maximum carrying capacity, we need to differentiate this expression with respect to h, set it equal to zero, and solve for h. Let's do that:
d/dh (8h - 2h^2) = 8 - 4h = 0
8 = 4h
h = 2
Now we have the value of h, which is 2 inches. To find the width of the base, we can substitute this value back into the expression for the width: 8 - 2h = 8 - 4 = 4 inches.
Therefore, the dimensions of the cross-section that will maximize the carrying capacity of the trough are 2 inches (height) by 4 inches (width).
2. In order to determine the size of the squares that should be cut to obtain a box with the largest possible volume, we can again use the concept of optimization.
Let's start by visualizing the sheet of cardboard and the squares that will be cut from the corners. Since the sheet of cardboard is 18 inches square, we can assume that x inches will be cut from each corner to form the open box.
Now, when we fold up the sides of the cardboard, the height of the resulting box will be x inches, and the length and width will be (18 - 2x) inches.
The volume of the box is determined by the length, width, and height, which is (18 - 2x) * (18 - 2x) * x = x(18 - 2x)^2 cubic inches.
To find the maximum volume, we need to differentiate this expression with respect to x, set it equal to zero, and solve for x. Let's do that:
d/dx (x(18 - 2x)^2) = (18 - 2x)^2 - 4x(18 - 2x) = 2(18 - 2x)(9 - x) - 4x(18 - 2x) = 4x^2 - 72x + 324 - 36x - 72x + 324 - 72x + 18x^2 = 22x^2 - 180x + 648 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 22, b = -180, and c = 648. Plugging these values into the formula, we find:
x = (-(-180) ± √((-180)^2 - 4 * 22 * 648)) / (2 * 22)
x = (180 ± √(32400 - 56928)) / 44
x = (180 ± √(-24528)) / 44
Since we're looking for a positive value for x, we can ignore the negative square root. Simplifying further, we find:
x ≈ 4.575 inches (rounded to three decimal places)
Therefore, to obtain a box with the largest possible volume, squares of approximately 4.575 inches should be cut from each corner of the cardboard.