0:00
Hello.
00:01
So from our given data, we observed the demand went up by 5 .5 pounds when the price goes down by 11 cents because the demand of chicken depends literally on the price.
00:14
So the decrease in price by one cent would increase the demand by 0 .5 pounds.
00:18
So here what we can do is we can let the price and demand can be expressed in terms of x.
00:25
If we let x be the number of cents and price per pound decreases, as we have that the price, then, p is going to be equal to 25 minus x, and then we can express the demand, q is equal to 22 plus 0 .5x.
00:40
So we know the revenue is the product of the price and quantity, so r is equal to p times q.
00:45
Therefore we get that r is going to be equal to, well, 25 minus x times 22 plus 0 .5x gives us that r is going to be equal to negative 0 .5x squared minus 9 .5x plus 550.
01:08
So to maximize this revenue, we find the derivative and then equate the derivative equal to zero.
01:13
So therefore, we find the derivative here, just differentiate.
01:16
Term by term, and we get r prime of x...