a) If the cost is raised to $160, n would be 2 (since $20 increase per person ÷ $10 increase per person = 2). Thus, substituting n=2 into the Revenue equation, R(2) = (140 + 20)(190) = $30,800. Therefore, the Student Council will make $30,800 in revenue if they raise the cost to $160/person.
b) To find R'(n), we first expand the R(n) equation:
R(n) = 140(200) + 10n(200) - 5n(140) - 50n^2
R(n) = 28,000 + 2,000n - 700n - 50n^2
R(n) = -50n^2 + 1,300n + 28,000
Taking the derivative of R(n) with respect to n:
R'(n) = -100n + 1300
Therefore, R'(n) = -100n + 1,300.
c) To find when the slope of the tangent to R(n) is zero, we set R'(n) = 0 and solve for n:
-100n + 1300 = 0
n = 13
Therefore, the slope of the tangent to R(n) is zero when n = 13.
d) The derivative of the Revenue function, R'(n) represents the rate of change of revenue with respect to the number of $10 increases in price. When R'(n) is positive, the revenue is increasing, and when R'(n) is negative, the revenue is decreasing. The maximum revenue occurs at the point where the derivative of the Revenue function is zero, or when the slope of the tangent to the Revenue function is zero. This is because at this point, the rate of change of revenue is neither increasing nor decreasing, indicating that the revenue is at a maximum. This is illustrated in the sketch below:
The blue line represents the Revenue function, and the orange line represents the derivative of the Revenue function. As n increases, the Revenue function initially increases at a decreasing rate, reaches a maximum at n=13, and then decreases at an increasing rate. The derivative of the Revenue function is positive before n=13 and negative after n=13. At n=13, the slope of the tangent to the Revenue function is zero, indicating that the maximum revenue is achieved at this point. Therefore, the derivative of the Revenue function and the maximum Revenue are related because the derivative helps to identify the point at which the maximum revenue is achieved.