Y varies directly with x. When x is 4 y is 8. Find y when x = 10.

A rectangular tank that is 2048 ft cubed with a square base and open top is to be constructed of sheet steel of a given thicknes

ira [324]

Answer:

The box should have base 16ft by 16ft and height 8ft Therefore,dimensions are 16 ft by 16 ft by 8 ft

Step-by-step explanation:

We were given the volume of the tank as, 2048 cubic feet.

Form minimum weight, the surface area must be minimum.

Let the height be h and the lengths be x

the volume will be: V=x²h then substitute the value of volume, we have

2048=hx²

hence

h=2048/x²

Since the amount of material used is directly proportional to the surface area, then the material needs to be minimized by minimizing the surface area.

The surface area of the box described is

A=x²+4xh

Then substitute h into the Area equation we have

A= x² + 4x(2048/x²)

A= x² + 8192/x

We want to minimize

A

dA/dx = -8192/x² + 2 x= 0 for max or min

when dA/dx=0

dA/dx= 2x-8192/x²=0

2x=8192/x²

Hence

2x³=8192

x³=4096

x=₃√(4096)

X=16ft

Then h=2048/x²

h=2048/16²

h=8ft

The box should have base 16ft by 16ft and height 8ft

Hence the dimensions are 16 ft by 16 ft by 8 ft

8 0

3 years ago

Help please:))

Blababa [14]

Answer:

a) Because this asks about the radius and height, I assume that we are talking about a cylinder shape.

Remember that for a cylinder of radius R and height H the volume is:

V = pi*R^2*H

And the surface will be:

S = 2*pi*R*H + pi*R^2

where pi = 3.14

Here we know that the volume is 1000cm^3, then:

1000cm^3 = pi*R^2*H

We can rewrite this as:

(1000cm^3)/pi = R^2*H

Now we can isolate H to get:

H = (1000cm^3)/(pi*R^2)

Replacing that in the surface equation, we get:

S = 2*pi*R*H + pi*R^2

S = 2*pi*R*(1000cm^3)/(pi*R^2) + pi*R^2

S = 2*(1000cm^3)/R + pi*R^2

So we want to minimize this.

Then we need to find the zeros of S'

S' = dS/dR = -(2000cm^3)/R^2 + 2*pi*R = 0

So we want to find R such that:

2*pi*R = (2000cm^3)/R^2

2*pi*R^3 = 2000cm^3

R^3 = (2000cm^3/2*3.14)

R = ∛(2000cm^3/2*3.14) = 6.83 cm

The radius that minimizes the surface is R = 6.83 cm

With the equation:

H = (1000cm^3)/(pi*R^2)

We can find the height:

H = (1000cm^3)/(3.14*(6.83 cm)^2) = 6.83 cm

(so the height is equal to the radius)

b) The surface equation is:

S = 2*pi*R*H + pi*R^2

replacing the values of H and R we get:

S = 2*3.14*(6.83 cm)*(6.83 cm) + 3.14*(6.83 cm)^2 = 439.43 cm^2

c) Because if we pack cylinders, there is a lot of space between the cylinders, so when you store it, there will be a lot of space that is not used and that can't be used for other things.

Similarly for transport problems, for that dead space, you would need more trucks to transport your ice cream packages.

3 0

3 years ago

3(b - 15) =9 solve for b

shepuryov [24]

B = 18 because 18-5 is equal to 3 and 3x3 is equal to 9.

4 0

3 years ago

In 2003 a town’s population was 1431. By 2007 the population had grown to 2134. Assume the population is changing linearly.How m

sergiy2304 [10]

We will have the following:

First, we will have that the population change by:

$\Delta P=2134-1431\Rightarrow\Delta P=703$

So, the population changed by 703 people.

It took the follwoing number of years to change from 1431 to 2134:

$\Delta y=2007-2003\Rightarrow\Delta y=4$

So, it took 4 years to changes from 1431 to 2134.

$m=\frac{2134-1431}{2007-2003}\Rightarrow m=175.75$

The average population growth per year is of 175.75 people per year.

The population in 2000 would have been:

$P_{2000}=\frac{2000\cdot1431}{2003}\Rightarrow P_{2000}=1428.856715\ldots$

$\Rightarrow P_{2000}\approx1429$

So, the population in 2000 would have been approximately 1429 people.

We will have that the equation for the population would be:

$P=(\frac{2134-1431}{2007-2003})t+1429\Rightarrow P=\frac{703}{4}t+1429$

So, the equation would be:

$P=175.75t+1429$

Now, we predict the number of people of the town in 2014:

$P_{2014}=175.75(14)+1429\Rightarrow P_{2014}=3889.5$

$\Rightarrow P_{2014}\approx3890$

So, there will be approximately 3890 people of the town in 2014.

3 0

1 year ago

100 POINTS!!! Orthographic projection and isometric projection are two ways to show three-dimensional objects in a two-dimension

shutvik [7]

Example 1:

The pros of Orthographic is that they can show hidden details and all of the connecting parts, they can be annotated to display material and finishes. The pros of Isometric projection is that they dont need many views and it gives accuracy, cons are is created a unorginized apperance by the lack of foreshortening, I would choose Isometric projection because it shows the size of the figure.

Example 2:

Orthographic projection is a good option for showing lots of detail and small things. The limitation is that with all of that detail, they can become quite messy and hard to understand to someone new to them. However, that is one of the pros of Isometric projection. It gives easy detail and is just as good as an Orthographic. Personally, I find Isometric projections easier to interpret.

4 0

3 years ago

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