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**mathxyz****Member**- From: Brooklyn, NY
- Registered: 2024-02-24
- Posts: 1,053

1. Imagine that you own a grove of orange trees, and suppose that from past experience you know that when 100 trees are planted, each tree will yield approximately 240 oranges per year. Furthermore, you’ve noticed that when additional trees are planted in the grove, the yield per tree decreases. Specifically, you have noted that the yield per tree decreases by about 20 oranges for each additional tree planted.

(a) Let y denote the yield per tree when x trees are planted. Find a linear equation relating x and y.

(b) Use the equation in part (a) to determine how many trees should be planted to obtain a yield of 400 oranges per tree.

(c) If the grove contains 95 trees, what yield can you expect from each tree?

2. Show that the slope of the line passing through the two points (a, a^2) and (x, x^2) is x + a.

3. Show that the slope of the line passing through the two points (3, 9) and (3 + h, (3 + h)^2) is 6 + h.

4. Show that the slope of the line passing through the two points (a, a^3) and (x, x^3) is x^2 + ax + a^2.

5. Write down, and then simplify as much as possible, an expression for the slope of the line passing through the two points (a, 1/a) and (x, 1/x).

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 10,588

1. We know that (100,240) is on the line. I had assumed to start with that no points exist before this but part B suggest we can extrapolate backwards.

If x increases by 1, y drops by 20 so the gradient is -20(x-100)

Putting these together we have

y = 240 - 20(x-100)

Check when x=101 we get y = 240 -20(101-100) = 240 - 20 = 220

When x = 99 y = 240 -20(99-100) = 240 + 20 = 260

2. Start with (a^2-x^2)/(a-x) and simplify.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**mathxyz****Member**- From: Brooklyn, NY
- Registered: 2024-02-24
- Posts: 1,053

Bob wrote:

1. We know that (100,240) is on the line. I had assumed to start with that no points exist before this but part B suggest we can extrapolate backwards.

If x increases by 1, y drops by 20 so the gradient is -20(x-100)

Putting these together we have

y = 240 - 20(x-100)

Check when x=101 we get y = 240 -20(101-100) = 240 - 20 = 220

When x = 99 y = 240 -20(99-100) = 240 + 20 = 260

2. Start with (a^2-x^2)/(a-x) and simplify.

Bob

Problems that I post here are not those for which I seek help with.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 10,588

Sorry again, my mistake. But I'm thinking this is an odd model.

y = 240 - 20(x-100)

So if the grower plants more trees the yield per tree drops (I'm thinking increased competition for resources).

But the model appears to have backward extrapolation: ie. we can set x to smaller values than 100, and get a bigger yield.

So what about x=1 ? y = 240 -20 times -99 = 240 + 1980 = 2220. So if we have just one tree we get 2220 oranges on it. Hhhmmm. I think the model needs to have a lower limit for the domain of x.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**mathxyz****Member**- From: Brooklyn, NY
- Registered: 2024-02-24
- Posts: 1,053

Bob wrote:

Sorry again, my mistake. But I'm thinking this is an odd model.

y = 240 - 20(x-100)

So if the grower plants more trees the yield per tree drops (I'm thinking increased competition for resources).

But the model appears to have backward extrapolation: ie. we can set x to smaller values than 100, and get a bigger yield.

So what about x=1 ? y = 240 -20 times -99 = 240 + 1980 = 2220. So if we have just one tree we get 2220 oranges on it. Hhhmmm. I think the model needs to have a lower limit for the domain of x.

Bob

I will study this problem a bit further.

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