1015

 

If we shave a cube so that its dimensions become .8 times as great, then

This gives us .8 * (.8 * ( .8 * original volume) ) ) = .8^3 * original volume = .512 times original volume.

How much salad can we get on a salad plate 8 inches in diameter, as opposed to a plate 10 inches in diameter?

Assume that anything we add to salad avalanches off when the sides reach a certain angle, which doesn't depend on the diameter of the plate.  Then the largest piles possible on both plates are geometrically similar cones.

In this case we can think of each salad as being made up of little cubes.  If we scale the salad on the larger plate down to the dimensions of the salad on the larger plate, the dimensions of the smaller cube all become .8 times as great.  So the smaller salad will be made up of an identical number of cubes, each with .8^3 = .512 times the volume of the larger.

So the smaller salad will contain only .512, a little more than half, as much salad as the larger, even though it has .8 times the diameter of the larger.

How much dressing will it take to cover the smaller salad to a depth of 1/4 inch over its surface, compared to the amount needed to cover the larger?

The surface of the larger salad can be thought of as being covered by tiny squares.  When the squares are scaled down to .8 times their original dimensions we end up with .8 * .8 * original surface area = .64 * original surface area.  So we will have to cover only .64 as much surface area on the smaller salad.

Note that we have .64 times as much dressing on the smaller salad, but only .512 times as much salad.  So there is more dressing for the amount of salad on the smaller.

 

On your test the question had to so with a 3-dimensional object whose dimension changed from 4 m to 8.7 m. 

 

The mass of the object is contained in its volume, and its volume is proportional to the cube of its linear dimension so we would have y = k x^3, or for your specific example

 

mass = k * diameter^3.

 

Given mass 1.4 million kg when diameter is 4 m we get

 

k = mass / diameter^3 = 1.4 million kg / (4 m)^3 = 21,000 kg / m^3

 

so that

 

mass = 21,000 kg / m^3 * diameter^3.

 

So when diameter is 8.7 m we have

 

mass = 21,000 kg / m^3 * (8.7 m)^3 = 14 million kg, approx.

 

This is consistent with your answer to the question.

 

To paint the surface we need only cover the surface, which can be thought of as being covered with little squares.  Surface area therefore scales up as the square of the linear dimension and we have

 

y = k x^2 or, using meaningful variable names,

paint_amt = k * diameter^2.

 

From the given information

 

k = paint_amt / diameter^2 = 8 liters / (4 m)^2 = .125 liters / m^2 so that

 

paint_amt = .125 liters / m^2 * diameter^2 and when diameter is 8.7 m we have

paint_amt = .125 liters / m^2 * (8.7 m)^2 = 20 liters, approximately.

 

Let me know if this is helpful.

 

Where are we?

 

Study of functions.

Functions can model important real-world things

This course is centered around a few types of basic functions

We understand functions in the following ways:

 

 

Topics we've covered or are in the process of covering:

 

Modeling 

The Basic Functions

 

Graphs

 

 

Equations

 

Properties of Quadratic Functions

Properties of Linear Functions

Properties of Power Functions