at a certain college the economics club plans to sell tshirts as a funder raiser. the president of the club estimated when the p

Question

3 years ago

13

at a certain college the economics club plans to sell tshirts as a funder raiser. the president of the club estimated when the p

rice of T-shirt is $18, there will be 60T shirts sold, and for every $1 the price of t-shirt is reduced, 10 additional T-shirts will be sold. Based on the estimate, what is the maximum amount of revenue, in dollars, the club can earn fro T-shirt sales? (Revenue equals price times the number sold. Disregard the $ sign when gridding your answer)

SAT

2 answers:

LenaWriter [7] · Answer 1 · 2021-12-06T15:39:52+03:00

Answer:

19,950 dollars

Explanation:

shirt no cost revenue

60 18 1080

70 17 1190

80 16 1280

90 15 1350

100 14 1400

110 13 1430

120 12 1440

130 11 1430

140 10 1400

150 9 1350

160 8 1280

170 7 1190

180 6 1080

190 5 950

200 4 800

210 3 630

220 2 440

230 1 230

19950

Though they're going to make a loss with over 1000 t shirts being sold at less than $5

alexdok [17] · Answer 2 · 2021-12-06T17:54:14+03:00

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