Answer:
$1,440
Explanation:
From the problem statement we can note the following:
1. At the time when a T-shirt costs $18, 60 of it will be sold.
2. For every $1 the price of t-shirt is reduced, 10 additional T-shirts will be sold.
Therefore, if we assume the price is dropped by x (dollars) from $18 to 18-x, then the quantity sold will be increased from 60 to 60+10x.
We can then formulate an expression for the revenue generated from selling the Tshirts as a function of the product of the price, 18-x, and the quantity 60+10x as follows:
R=(18-x)(60+10x) (1)
We solve for the optimum values of this quadratic function, R, to determine its maximum.
Expanding the right hand side, we have
R= 1080-60x+180x-10x^2 = 1080+120x-10x^2 (2)
To determine the optimum value, we equate the first differential, dR/dx = 0,
dR/dx=120-20x=0
120 = 20x
x=6
to ascertain the nature of this optimum, we can find the second differential
d2R/dx2=-20<0 , thus we must have a maximum value at x=6.
finally, we evaluate the maximum value by inserting x=6 into equation (1):
R max = (18-6)(60+10*6)=12*120=1440
the maximum revenue will therefore be $1440
