House of Mohammed sells packaged lunches, where their finance department has established a
weekly relationship between its Revenue R, in dollars and the number of lunches x, as R=x(82−x)
(a) Write the revenue R, in the form R=ax2+bx+c (b) What is the revenue when 35 lunches are sold? (c) Explain why the graph of R has a maximum.
(d) Write R in the form a(x−h)2+k
(e) How many lunches must be sold for them to achieve their maximum revenue?
(f) State the company's maximum revenue.
(2 marks) (2 marks) (2 marks)
(5 marks)
(2 marks) (2 marks)
(g) Complete the following table of values and graph the Revenue equation using the ordered pairs
(x , R). Label the coordinates of the vertex.
x
10
20
30
40
50
60
70
80
90
R
( 10 marks)

Answers

Answer 1

The revenue function is a quadratic equation and the graph of the function

has the shape of a parabola that is concave downwards.

The correct responses are;

(a) R = -x² + 82·x(b) $1,645(c) The graph of R has a maximum because the leading coefficient of the quadratic function for R is negative.(d)  R = -1·(x - 41)² + 1,681(e) 41(f) $1,681

Reasons:

The given function that gives the weekly revenue is; R = x·(82 - x)

Where;

R = The revenue in dollars

x = The number of lunches

(a) The revenue can be written in the form R = a·x² + b·x + c by expansion of the given function as follows;

R = x·(82 - x) = 82·x - x²

Which gives;

R = -x² + 82·x

Where, the constant term, c = 0

(b) When 35 launches are sold, we have;

x = 35

Which by plugging in the value of x = 35, gives;

R = 35 × (82 - 35) = 1,645

The revenue when 35 lunches are sold, R = $1,645

(c) The given function for R is R = x·(82 - x) = -x² + 82·x

Given that the leading coefficient is negative, the shape of graph of the

function R is concave downward, and therefore, the graph has only a

maximum point.

(d) The form a·(x - h)² + k is the vertex form of quadratic equation, where;

(h, k) = The vertex of the equation

a = The leading coefficient

The function, R = x·(82 - x), can be expressed in the form a·(x - h)² + k, as follows;

R = x·(82 - x) = -x² + 82·x

At the vertex, of the equation; f(x) = a·x² + b·x + c,  we have;

[tex]\displaystyle x = \mathbf{-\frac{b}{2 \cdot a}}[/tex]

Therefore, for the revenue function, the x-value of the vertex, is; [tex]\displaystyle x = -\frac{82}{2 \times (-1)} = \mathbf{41}[/tex]

The revenue at the vertex is; [tex]R_{max}[/tex] = 41×(82 - 41) = 1,681

Which gives;

(h, k) = (41, 1,681)

a = -1 (The coefficient of x² in -x² + 82·x)

The revenue equation in the form, a·(x - h)² + k is; R = -1·(x - 41)² + 1,681

(e) The number of lunches that must be sold to achieve the maximum revenue is given by the x-value at the vertex, which is; x = 41

Therefore;

The number of lunches that must be sold for the maximum revenue to be achieved is 41 lunches

(f) The maximum revenue is given by the revenue at the vertex point where x = 41, which is; R = $1,681

The maximum revenue of the company is $1,681

Learn more about the quadratic function here:

https://brainly.com/question/2814100


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[tex]y = 2(x+4)^2+5\implies y = 2(x+4)(x+4)+5 \\\\\\ y = 2(\stackrel{\mathbb{USING~FOIL}}{x^2+8x+16})+5\implies y= 2x^2+16x+32+5 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y = 2x^2+16x+37~\hfill[/tex]

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