A concert promoter needs to make $43,600 from the sale of 1880 tickets. The promoter charges $20 for some tickets and $30 for the others. Let x represent the number of $20 tickets and y represent the number of $30 tickets.

(a) Write an equation that states that the sum of the tickets sold is 1880.

(b) Write a formula for how much money is received from the sale of $20 tickets?

$

(c) Write a formula for how much money is received from the sale of $30 tickets?

$

(d) Write an equation that states that the total amount received from the sale is $43,600.

(e) Solve the equations simultaneously to find how many tickets of each type must be sold to yield the $43,600.

x =

y =

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.