Function word problems

**archistrategos214** · January 21st 2007, 03:48 PM

Hi, I need help on a couple of problems. I'll be having my math mid-terms on the 26th, which is a Friday, and we were told by the teacher that some of the questions appearing in the test would be similar to the problems below. Specifically, I need help on four of them.

1) 1000 copies of a souvenir program will be sold if the price is $50 and that the number of copies sold decreases by 10 for each $1 added to the price. Write a function that determines gross revenue as a function of x. What price yields the largest gross sales revenue and what is the largest gross revenue? How many copies must be sold to yield the maximum gross revenue?

2) A rectangular field is fenced along a river bank, which is not. The material for the fence costs $12 per foot for the side parallel to the river, and while it costs $8 for the two other sides. What are the dimensions of the largest possible rec. field that can be enclosed with a budget of $3600? assume all material would be used up.

3) A waterway whose cross-section is a parabola is being designed for a theme park boat ride. It has a depth of 12ft. and a width of 6ft. at the surface. Rectangular cross-sectioned boats are to be used, and they sink to a depth of 4ft. What should be the proper width of the boat to facilitate easy passage?

[Note: parabola opens upwards]

4) Describe each of the equations in the given non-linear system.

a) (x-3)^2 = 9
-- y + 9
2

b) (x-3)^2 t y^2 = 4

**CaptainBlack** · January 21st 2007, 07:12 PM

Originally Posted by archistrategos214

1) 1000 copies of a souvenir program will be sold if the price is $50 and that the number of copies sold decreases by 10 for each $1 added to the price. Write a function that determines gross revenue as a function of x. What price yields the largest gross sales revenue and what is the largest gross revenue? How many copies must be sold to yield the maximum gross revenue?

The sales at price $x$ are given by the equation:

$S(x)=1000-10(x-50)$ ,

and the gross revenue then is:

$R(x)=x\, S(x)=1500\, x-10\,x^2=x(1500-10\,x$

Thus the graph of revenue against price is a parabola that opens downwards, and has its maximum midway between its roots. The roots of the quadratic are $x=0$ , and $x=150$ , so the price that maximises revenue is $\$75$ .

Now plug the $\$75$ into the equation for revenue to get the maximum revenue:

$R(75)=75 (1500-750)=75\times 750=\$5,671,800$

RonL

**Soroban** · January 21st 2007, 07:49 PM

Hello, archistrategos214!

2) A rectangular field is fenced along a river bank, which is not.
The material for the fence costs $12 per foot for the side parallel to the river,
while it costs $8 per foot for the two other sides.
What are the dimensions of the largest possible field
that can be enclosed with a budget of $3600?

Code:

      - * - - - - - - - * -
        |               |
       y|               |y
        |               |
        *---------------*
                x

Let $x$ = length of side parallel to the river.
. . At $12/ft, this will cost: $12x$ dollars.

Let $y$ = length of the other two sides.
. . At $8/ft, they will cost: $8(2y) = 16y$ dollars.

Using all the budget: . $12x + 16y \:=\:3600\quad\Rightarrow\quad y \:=\:225 - \frac{3}{4}x$ [1]

The area of the field is: . $A \;=\;xy$ [2]

Substitute [1] into [2]: . $A \;=\;x\left(225 - \frac{3}{4}x\right) \;=\;225x - \frac{3}{4}x^2$

Differentiate and equate to zero:
. . $A' \;=\;225 - \frac{3}{2}x\:=\:0\quad\Rightarrow\quad\boxed{ x\,=\,150\text{ ft}}$

Substitute into [1]: . $y \:=\:225 - \frac{3}{4}(150)\quad\Rightarrow\quad\boxed{ y\,=\,112.5\text{ ft}}$

**archistrategos214** · January 22nd 2007, 05:17 PM

Thanks for the help, guys! It is really appreciated.

Does anyone have any tips for number three? I tried solving it, but came out unsuccessful. Any help is most welcome!

**Soroban** · January 22nd 2007, 06:47 PM

Hello again, archistrategos214!

3) A waterway whose cross-section is a parabola is being designed for a theme park boat ride.
It has a depth of 12 ft and a width of 6 ft at the surface.
Rectangular cross-sectioned boats are to be used, and they sink to a depth of 4 ft.
What should be the proper width of the boat to facilitate easy passage?

Code:

                    |
  * - -*------------+--------------*- - *(3,12)
       |            |              |
   *   |            |             4|   *
    *  |     x      |      x       |  *
       *------------+--------------*
          *         |           *  :y
              *     |     *        : 
- - - - - - - - - - * - - - - - - -+- - - 
                    |

The parabola has the equation: . $y \,=\,ax^2$

The point $(3,12)$ is on the parabola.
. . We have: . $12 \:=\:a\cdot3^2\quad\Rightarrow\quad a = 2$
The parabola is: . $y \,=\,2x^2$

Since the boats sink to 4 feet, then: . $12 - y = 4$
. . Then: . $12 - 2x^2\:=\:4\quad\Rightarrow\quad x \,=\,2$

Therefore, the boatrs should be $2x = 4$ feet wide at most.

**archistrategos214** · January 30th 2007, 03:10 AM

"Given the equation 8y = x^2 + *x + 32 represents a parabola with vertical axis of symmetry; find the vertex, focus, directrix, axis of symmetry and x and y intercepts."

Just to double check. I solved this one and came up with a different graph than a parabola.

**topsquark** · January 30th 2007, 03:30 AM

Originally Posted by archistrategos214

"Given the equation 8y = x^2 + *x + 32 represents a parabola with vertical axis of symmetry; find the vertex, focus, directrix, axis of symmetry and x and y intercepts."

Just to double check. I solved this one and came up with a different graph than a parabola.

Three comments:
1) You should put new questions in a new thread.

2) There's a typo in your equation.

3) If the x term in the equation has a constant coefficient, then you did something wrong when you graphed it.

-Dan

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