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.
A. Identify the input quantity of the function.
B. Identify the output quantity of the function.
C. 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.)
D. Re-write your formula in y = mx + b form. Show your work.
E. Identify a Practical Domain for the function and explain your reasoning for the choice.
F. Identify a Practical Range for the function and explain your reasoning for the choice.
G. Explain the meaning of g- (12) in the context of the problem.
H. Determine the inverse formula for P = g(c).
1. Evaluate g-' (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].
  • The inverse function is: [tex]g^{-1}(c) = \frac{1}{22.9}(c - 7)[/tex] .
  • A practical domain is from 0% to 100%
  • A practical range is from 7 to 29.9

A. Input quantity

The input quantity is the percentage of frozen citrus crop

B. Output quantity

The output quantity is the cost of box of oranges

C. The linear function

We have:

[tex](P_1,c_1) = (20\%,11.58)\\(P_2,c_2) = (80\%,25.32)[/tex]

Calculate the slope of the function

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

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

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

[tex]m = 22.9[/tex]

The linear equation is calculated as follows:

[tex]c -c_1 = m(P-P_1)[/tex]

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

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

D. Rewrite as y = mx + b

We have:

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

Collect like terms

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

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

The function is:

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

E. A practical domain

The domain is the possible values of P.  Because P is a percentage, its possible values are 0% to 100%.

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

F. A practical range

When P = 0%

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

When P = 100%

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

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

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

The inverse function of g(P) is [tex]g^{-1}(P)[/tex]

So:

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

H. The inverse formula

We have:

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

Subtract 7 from both sides

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

Make P the subject

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

So, the inverse formula is:

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

Substitute 12 for c

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

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

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

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