A devastating freeze in California’s Central Valley in January 2007 wiped out approximately 75% of the state’s citrus crop. It turns out that the cost for a box of oranges is a function of the percentage of the citrus crop that was frozen, i.e. c=g(P) , where c is the price of a box of oranges and P is the percentage of the citrus crop that was frozen. When only 20% of the crop was frozen, the price for a box of oranges was $11.58. However, the price per box was $25.32 when 80% of the crop was frozen.



Identify the input quantity of the function.
Identify the output quantity of the function.
Using the method demonstrated in the Module 3 presentation, create a formula for a linear function that represents this situation. i.e. your formula should use one of the given reference points rather than the vertical intercept. (Note: You will need to first find the constant rate of change.)
Identify a Practical Domain for the function and explain your reasoning for the choice.
Identify a Practical Range for the function and explain your reasoning for the choice.
Explain the meaning of g−1(12) in the context of the problem.
Determine the inverse formula for P=g−1(c) .
Evaluate g−1(12) .

Respuesta :

The relationship between the percentage of frozen citrus crop, and the cost of box of oranges is an illustration of a linear function.

  • The linear equation of the function is: [tex]g(P) =22.9P +7[/tex], while its inverse is: [tex]g^{-1}(c) = \frac{c - 7}{22.9}[/tex].
  • The domain is from 0% to 100%, while the range is from 7 to 29.9

Given that:

[tex]c = g(P)[/tex]

Input and Output quantity

The input quantity is the percentage of frozen citrus crop, while the output quantity is the cost of box of oranges

Linear Function

The given parameters can be written as:

[tex](P_1,c_1) = (20\%, 11.58)[/tex]

[tex](P_2,c_2) = (80\%, 25.32)[/tex]

Calculate the slope (m)

[tex]m = \frac{c_2 - c_1}{P_2 - P_1}[/tex]

So, we have:

[tex]m = \frac{25.32 - 11.58}{80\% - 20\%}[/tex]

[tex]m = \frac{13.74}{60\%}[/tex]

[tex]m = 22.9[/tex]

The equation is then calculated using:

[tex]c =m(P - p_1) + c_1[/tex]

So, we have:

[tex]c =22.9(P - 20\%) + 11.58[/tex]

[tex]c =22.9P - 4.58 + 11.58[/tex]

[tex]c =22.9P +7[/tex]

So, the function is:

[tex]g(P) =22.9P +7[/tex]

The domain and the range

The domain is the possible input value (i.e. possible values of P).

Because P is a percentage, its possible values are 0% to 100%.

Hence, the domain of the function is: [tex][0\%,100\%][/tex]

The range is the possible output value (i.e. possible values of c).

When P = 0% and 100%

[tex]c = 22.9 \times 0\% + 7 = 7[/tex]

[tex]c = 22.9 \times 100\% + 7 = 29.9[/tex]

Hence, the range of the function is: [tex][7,29.9][/tex]

The meaning of [tex]g^{-1}(12)[/tex]

[tex]g^{-1}(12)[/tex] is an inverse equation, where 12 represents the cost of box of oranges.

So, [tex]g^{-1}(12)[/tex] represents the percentage of frozen citrus crop, when the cost is $12.

The inverse formula

We have:

[tex]c =22.9P +7[/tex]

Make P the subject

[tex]22.9P = c - 7[/tex]

Divide by 22.9

[tex]P = \frac{c - 7}{22.9}[/tex]

So, the inverse function is:

[tex]g^{-1}(c) = \frac{c - 7}{22.9}[/tex]

This is used to calculate the percentage of frozen citrus crop, when the cost is known.

[tex]g^{-1}(c) = \frac{c - 7}{22.9}[/tex]

Substitute 12 for c

[tex]g^{-1}(12) = \frac{12 - 7}{22.9}[/tex]

[tex]g^{-1}(12) = \frac{5}{22.9}[/tex]

[tex]g^{-1}(12) = 22\%[/tex]

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