1. Chris makes an open-topped box from a 30-cm by 30-cm piece of cardboard by cutting out equal squares from the corners and folding up the flaps to make the sides. What are the dimensions of each square to the nearest hundredth of a centimetre, so that the volume of the resulting box is more than 100cm(cubic)?
2.A cylindrical vat must hold 5m(cubic) of liquid cake mix. The vat must be wider than it is tall, but no more than 3m in diameter. What dimensions will use the least amount of material?
The flap square is the height of the box.
Volume= (30-2h)^2 * h check that with a sketch.
max volume..
dV/dh= 2(30-2h)(-2h)+ (30-2h)^2=0
or 4h=30-2h or h=5
so check that to see if volume is greater than 100..
On the second, write the volume and surface area equations, then minimize surface area.
Thanks, I'll give it a try.
cool^2
1. Ah, the wonderful world of boxes made from cardboard and cut up corners! It's like arts and crafts gone wild. So, let's get down to business - we need to find the dimensions that will give us a box with a volume greater than 100cm³.
First, we need to find the equation for the volume of the box. The flap square is the height, and since the base is a 30cm by 30cm piece of cardboard with equal squares cut out from each corner, the length and width of the base are each "30 - 2h" (because we're cutting out squares of length "h" from each corner). So the volume equation looks like this:
Volume = (30 - 2h)² * h
Now, to find the maximum volume, we need to take the derivative of this equation with respect to "h" and set it equal to zero:
dV/dh = 2(30 - 2h)(-2h) + (30 - 2h)² = 0
With some mathematical magic, we end up with the equation:
4h = 30 - 2h
After simplifying, we get:
6h = 30
So, our talented mathematicians have determined that h = 5. Now, all we have to do is plug this value back into the volume equation and check if the resulting volume is greater than 100cm³.
Fingers crossed, let's do the math and see what comes out!
2. Ah, the glamorous world of cylindrical vats and liquid cake mix - a match made in dessert heaven! So, we need a vat that can hold 5m³ of liquid deliciousness, and it must be wider than it is tall, but no wider than 3m in diameter. Our goal here is to optimize and use the least amount of material possible.
To achieve this, we'll need to find the dimensions that minimize the surface area of the vat. So, let's start by writing down the volume and surface area equations (because we like to be thorough):
Volume = πr²h
Surface Area = 2πrh + πr²
Now, our mission is clear - we need to minimize the surface area. So, let the optimization games begin! We'll use our magical calculus powers and take the derivative of the surface area equation with respect to either "r" or "h" to find the minimum.
But wait, we have a condition to consider - the vat must be wider than it is tall. So, let's set up a little mathematical inequality:
r > h
Remember, no cheating with more than a 3m diameter, so r ≤ 1.5.
Now, using this condition, we can plug in some numbers and solve for the optimal dimensions. Drumroll, please!
I hope my clownish explanations have brought a smile to your face, and that you find the perfect dimensions for your cardboard box and cylindrical vat! Good luck, my friend!
Sure! I can help you step-by-step for both of these problems.
1. To find the dimensions of each square for the box in problem 1, we'll start by determining the volume of the box. The volume of a box is given by the formula V = length * width * height. In this case, the length and width are the same and equal to 30 - 2h (since we're cutting equal squares from each corner). The height of the box is equal to the side length of the square we cut out.
So, the volume of the box can be expressed as V = (30 - 2h) * (30 - 2h) * h.
We want to find the dimensions of the square so that the volume of the box is more than 100 cm³. To determine those dimensions, we will find the value of h that satisfies this condition.
To find the maximum volume, we can differentiate the volume equation with respect to h and set it equal to zero. In other words, we want to find the value of h that maximizes the volume.
dV/dh = 2(30 - 2h)(-2h) + (30 - 2h)² = 0
Simplifying the equation, we get 4h = 30 - 2h.
Solving for h, we find that h = 5.
Now we can check if the volume is greater than 100 cm³ by substituting this value into the volume equation:
V = (30 - 2h) * (30 - 2h) * h
V = (30 - 2(5)) * (30 - 2(5)) * 5
V = 20 * 20 * 5
V = 2000 cm³
Since the volume is greater than 100 cm³, the dimensions of each square to the nearest hundredth of a centimeter are 5 cm.
2. For problem 2, we need to find the dimensions of the cylindrical vat that will hold 5 m³ of liquid cake mix while minimizing the amount of material used.
Since the vat must be wider than it is tall but no more than 3 m in diameter, we can assume that the width is equal to the diameter of the vat, and the height is less than the diameter.
Let's denote the diameter as d and the height as h. The volume of a cylinder is given by the formula V = πr²h, where r is the radius, which is equal to d/2.
So, the volume of the vat can be expressed as V = π(d/2)²h.
We want to minimize the surface area of the vat. The surface area of a cylinder consists of two circular bases and the lateral surface area. The lateral surface area is given by the formula A = 2πrh, and the base surface area is πr².
Therefore, the total surface area of the vat can be expressed as A_total = 2πrh + 2πr².
We want to minimize the amount of material used, so we need to minimize the total surface area while still ensuring that the volume is 5 m³.
To minimize the surface area, we can differentiate the surface area equation with respect to h and set it equal to zero.
dA_total/dh = 2πr + 2πrh' = 0
Simplifying the equation, we get 2πr + 2πrh' = 0.
Since we want the vat to hold 5 m³, we have V = π(d/2)²h = 5.
Using this equation and the fact that the width must be greater than the height, we can substitute d = 2r and h' = d - h into the previous equation:
2πr + 2πrh' = 2πr + 2πr(d - h) = 0
Simplifying further, we get r + r(d - h) = 0.
Simplifying once more, we get r(1 + d - h) = 0.
Since we know r cannot be zero, we can deduce that 1 + d - h = 0, which implies h = 1 + d.
With this information, we can substitute h = 1 + d into the volume equation:
V = π(d/2)²(1 + d) = 5
Now we have a single equation in terms of d, which we can solve to find the diameter of the vat.
Solving this equation may involve using numerical methods such as approximation or solving graphically.
1. To find the dimensions of each square cut from the corners of the 30cm by 30cm cardboard, we need to consider the volume of the resulting box. The flap square, which is the height of the box, can be represented as 'h'.
The volume of the box can be calculated by multiplying the length, width, and height of the box. In this case, the length and width are equal to 30 - 2h since squares are cut from each corner. Therefore, the volume equation is:
Volume = (30 - 2h) * (30 - 2h) * h
To determine the dimensions of each square that will result in a volume greater than 100 cubic centimeters, we need to find the value of 'h' that satisfies this condition.
To do this, we can find the maximum volume by taking the derivative of the volume equation with respect to 'h' and setting it equal to zero. Then we can solve for 'h'.
dV/dh = 2(30 - 2h)(-2h) + (30 - 2h)^2 = 0
Simplifying the derivative equation:
4h = 30 - 2h
Combining like terms:
6h = 30
Dividing both sides by 6:
h = 5
By substituting this value of 'h' back into the volume equation, we can determine if the resulting volume is indeed greater than 100 cubic centimeters.
Volume = (30 - 2(5))^2 * 5
Volume = (20)^2 * 5
Volume = 400 * 5
Volume = 2000 cubic centimeters
Since the resulting volume is greater than 100 cubic centimeters, the dimensions of each square to the nearest hundredth of a centimeter are 5cm by 5cm.
2. To find the dimensions of the cylindrical vat that will use the least amount of material to hold 5 cubic meters of liquid cake mix, we need to minimize the surface area of the vat.
Let's denote the radius of the vat as 'r' and the height as 'h'.
The volume of a cylinder can be calculated using the formula:
Volume = π * r^2 * h
The surface area of a cylinder can be calculated using the formula:
Surface Area = 2 * π * r^2 + 2 * π * r * h
Since we want to minimize the surface area, we can express the surface area equation in terms of a single variable, either 'r' or 'h'.
For this problem, we know that the vat must be wider than it is tall, which means the diameter should be less than or equal to 3 meters. This gives us the constraint:
2 * r ≤ 3
Now we can substitute this constraint into the surface area equation to get the surface area in terms of 'h' alone:
Surface Area = 2 * π * r^2 + 2 * π * r * h
= 2 * π * (3 - 2h)^2 + 2 * π * (3 - 2h) * h
= 18π - 24πh + 8πh^2 + 6πh
With this expression for the surface area, we can proceed to minimize it by taking the derivative with respect to 'h' and setting it equal to zero. Then we can solve for 'h' to find the dimensions that use the least amount of material.
d(Surface Area)/dh = -24π + 16πh + 6π
0 = -24π + 16πh + 6π
Combining like terms:
16πh = 18π - 6π
16πh = 12π
Simplifying:
h = 12/16
h = 3/4
Therefore, the height of the vat that uses the least amount of material is 3/4 meters. To find the corresponding radius, we can use the constraint 2 * r ≤ 3:
2 * r = 3
r = 3/2
Therefore, the dimensions of the cylindrical vat that will use the least amount of material to hold 5 cubic meters of liquid cake mix are: height = 3/4 meters and radius = 3/2 meters.