# help! exam tomorrow

• Jun 21st 2007, 04:21 PM
eric1044
help! exam tomorrow
1)Two hundred people attended a dance recital. Adults tickets were $10.00 and children were$7.50. If the total revenue was $1850.00 how many adults and how many children attended? 2) Forty metres of fencing are available to enclose a rectangular garden. The area of the garden is given by the formula A= x(20-x) where x is the length of the garden. a) What is the maximum area that can be enclosed?? b) What are the dimensions of the garden with maximum area? If anyone knows how to do this please help, its much appreciated. • Jun 21st 2007, 04:28 PM eric1044 even if someone helps me with just one, that will help! • Jun 21st 2007, 06:01 PM Soroban Hello, Eric! Quote: 1)Two hundred people attended a dance recital. Adults tickets were$10.00 and children were $7.50. If the total revenue was$1850.00,
how many adults and how many children attended?

Let $A$ = number of adults.
Then $(200 - A)$ = number of children.

The $A$ adult's tickets brought $10 each. . . The revenue was: $10A$ dollars. The $(200-A)$ children's tickets brought$7.50 each.
. . The revenue was: $7.5(200 - A)$ dollars.
Hence, the total revenue was: . $10A + 7.5(200 - A)$ dollars.

But we are told that the total revenue was 1850 dollars.

There is our equation! . . . . $10A + 7.5(200 - A) \:=\:1850$

Solve for $A\!:\;\;10A + 1500 - 7.5A \:=\:1850\quad\Rightarrow\quad 2.5A \:=\:350$

Hence: . $A \,=\,140$

Then: . $200 - 140 \:=\:60$

There were: $\boxed{\text{140 adults and 60 children.}}$

Quote:

2) Forty metres of fencing are available to enclose a rectangular garden.
The area of the garden is given by the formula: . $A \:= \:x(20-x)$
. . where $x$ is the length of the garden.

a) What is the maximum area that can be enclosed?
b) What are the dimensions of the garden with maximum area?

The area is given by: . $A \:=\:20x - x^2$
This is a down-opening parabola; its vertex is the maximum point.

The vertex of a parabola is at: . $x \:=\:\frac{-b}{2a}$
We have: . $a = -1,\;b = 20$
So the vertex is at: . $x \:=\:\frac{-20}{2(-1)}\quad\Rightarrow\quad x\:=\:10$

(a) Then the maximum area is: . $A \:=\:20\!\cdot\!10 - 10^2 \:=\:\boxed{100\:m^2}$

Since the length is 10 m and the perimeter is 40 m, the width is also 10 m.

(b) The dimensions are: $\boxed{10 \times 10\text{ meters.}}$
. . .(The garden is square!)