Task B-2.8. A wholesale trade sells tennis rackets at a price of 363 kuna for one racket if the number of ordered rackets is 40 or less. If the number of ordered rackets is greater than 40, the price of one racket decreases by 0.60 kuna for each ordered racket above 40. In one order, a maximum of 400 rackets can be ordered. Determine how many rackets need to be ordered for the trade to achieve the maximum profit. How much is the maximum profit and at what price would the customer then pay for one racket?
Solution. Let $x$ be the number of rackets ordered.

The price per racket is $363-(x-40) \cdot 0.6$, or $387-0.6 x$.

(2 points)

The profit $d(x)$ that the store will achieve depends on the number of rackets ordered, and is calculated by the formula

$$
d(x)=x \cdot(387-0.6 x)=-0.6 x^{2}+387 x
$$

Since this is a quadratic function that has a maximum value for

$$
x_{\text {max }}=-\frac{b}{2 a}=\frac{387}{1.2}=322.5
$$

and since $322.5 \in[41,400]$, or within the domain of the function $d$, the maximum value of the function should be for $x_{\text {max }}=322.5$.

(2 points)

However, since $x$ must be an integer, the number of rackets ordered that brings the highest profit to the store is 322 or 323. Due to the symmetry of the graph of the quadratic function, both numbers give the same value of the profit function:

$$
d(322)=d(323)=62403.6 \text{ kn}
$$

The price of the rackets in this case is 193.8 kn or 193.2 kn.

Note: If a student writes only one solution for the price and the number of rackets, they lose 2 points, and if they do not round the number of rackets to an integer, they lose 3 points.

## COUNTY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B variant

March 13, 2012.

IF A STUDENT HAS A DIFFERENT METHOD OF SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT METHOD APPROPRIATELY.