maths coursework for tommorrow

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October 30th 2006, 12:20 PM
rom

maths coursework for tommorrow

Hi Guys, New to the place, its nice isn't it...:) comfy!!... anyway just need a helping hand with the question below. i have it for coursework and i am stuck!! please help cheers!:)

The manufacturer of an electronic gadget has weekly production costs given by C=3600+100x+2x[squared] . Also, there is weekly revenue from sales of the product given by R=(500-2x)x[squared] . In both cases x denotes the number of gadgets. The producer reaches a break-even point when production costs equal the revenue, otherwise either a loss or a profit is being made. Find the number of units for which a break-even point is reached and the number of units for which maximum revenue is obtained. At this maximum, is the producer losing money or making a profit?

please help!! i haven't got a clue where to start!
October 30th 2006, 01:29 PM
Soroban

Hello, rom!

Welcome aboard!
I assume (I hope) that this is from a Calculus course.

Quote:

A manufacturer has weekly production costs given by: . $C\:=\:3600+100x+2x^2$
Also, there is weekly revenue from sales of the product given by: . $R \:=\:(500-2x)x^2$
In both cases $x$ denotes the number of gadgets.

The producer reaches a break-even point when production costs equal the revenue,
otherwise either a loss or a profit is being made.

Find (a) the number of units for which a break-even point is reached
. . and (b) the number of units for which maximum revenue is obtained.
At this maximum, is the producer losing money or making a profit?

$\text{Profit }\;=\;\text{ Revenue } - \text{ Cost}$

$P(x) \;=\;(500 - 2x)x^2 - (3600 + 100x + 2x^2) \;=\;-2x^3 + 498x^2 - 100x - 3600$

(a) The break-even point is reached when $P(x) = 0.$
I don't know how they expect us to solve the cubic:
. . $-2x^2 + 498x^2 - 100x - 3600 \:=\:0$

(b) For maximum profit, differentiate $P(x)$ and equate to zero.
. . $P'(x)\:=\:-6x^2 + 996x - 100 \:=\:0$

This quadratic has roots: . $x \:=\:\frac{498 \pm \sqrt{247,404}}{6} \:\approx \:0.1,\;166$

Maximum profit is: . $P(166) \:=\:\$4,554,096$
October 30th 2006, 01:58 PM
rom

That Was Great!!! but Theres One More!!!!!! help!!

thank you very much that has helped hugely!! i do have another that i need help with if your not too busy......as you were so helpful in the first place!!

A farmer has 6000ha available to plant with corn and wheat. Each hectare of corn requires 9Litres of fertilizer and 3/4hr of labour to harvest. Each hectare of wheat requires 3Litres of fertilizer and 1hr labour to harvest. The farmer only has available at most 40,500ltrs of fertilizer and at most 5250h of labour for harvesting. If x represents the number of hectares planted with corn, and y represents the number of hectares planted with wheat, write the system of linear inequalities that describes the constraints and graph the feasible region for the system. The profits per hectare are £60 for corn and £40 for wheat. Write and equation for p , the total profit. How many hectares of each crop should the farmer plant in order to maximise profit? What is the maximum profit?

any help would really be appreciated, yeah i am doing a calculus module..
October 30th 2006, 02:01 PM
topsquark

Quote:

Originally Posted by Soroban

(a) The break-even point is reached when $P(x) = 0.$
I don't know how they expect us to solve the cubic:
. . $-2x^2 + 498x^2 - 100x - 3600 \:=\:0$

Well, ThePerfectHacker won't like it but you could easily use a program to approximate the solutions. Or we could use Cardano's method. (I shudder to even think of it, though!)

I get positive solutions at approximately x = 2.8074 and x = 248.76992. As I doubt the manufacturer would settle for only 2 units a week, I'd say it should be 248.

-Dan
October 30th 2006, 02:04 PM
rom

so is the producer making a profit?
October 30th 2006, 06:26 PM
Soroban

Hello again, rom!

Quote:

A farmer has 6000 ha available to plant with corn and wheat.
Each hectare of corn requires 9 L of fertilizer and 3/4 hr of labour to harvest.
Each hectare of wheat requires 3 L of fertilizer and 1 hr labour to harvest.
The farmer only has available at most 40,500 L of fertilizer
and at most 5250 h of labour for harvesting.

If $x$ represents the number of hectares planted with corn,
and $y$ represents the number of hectares planted with wheat,
(a) write the system of linear inequalities that describes the constraints
and graph the feasible region for the system.

The profits per hectare are £60 for corn and £40 for wheat.
(b) Write an equation for $P$ , the total profit.

(c) How many hectares of each crop should be planted to maximise profit?

(d) What is the maximum profit?

This is a problem in Linear Programming ... no calculus required.

Part (a)

The $x$ units of corn need $9x$ liters of fertilizer.
The $y$ units of wheat need $3y$ liters of fertilizer.
The fertilizer is limited to 40,500 L.
. . So we have: . $9x + 3y\:\leq \:40,500\quad\Rightarrow\quad\boxed{3x + y \:\leq \:13,500}$

The $x$ units of corn need $\frac{3}{4}x$ hours to harvest.
The $y$ units of wheat need $1y$ hours to harvest.
The total harvest time is limit to 520 hours.
. . So we have: . $\boxed{\frac{3}{4}x + y \:\leq \:5250}$

Code:

| 13,500* |* | * | * | * 5250 o * |:::* * |::::::o |:::::::* * |::::::::* * --o---------o------*-- - | 4500 7000

The graph is the region in Quadrant 1, below the two slanted lines.

. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(b) The profit is 60 per $x$ and 40 per $y.$

. . . Therefore: . $\boxed{P \:=\:60x + 40y}$

. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(c) For maximum profit, we consider only the vertices of the shaded region.

We know three vertices: . $(4500,0),\;(0,5250)$ , and $(0,0)$ (which we'll disregard).

The intersection of $3x + y \:=\:13,500$ and $3x + 4y \:=\:21,000$
. . is: . $\left(\frac{11,000}{3},\,2500\right)$ ... the fourth vertex.

$\begin{array}{ccc}(4500,\,0): \\ (0,\,5250): \\ \left(\frac{11,000}{3},\,2500\right):\end{array}<br /> \begin{array}{ccc}P \:= \\ P\;= \\ P\:=\end{array}$ $\begin{array}{ccc} 60(4500) + 40(0) \\ 60(0) + 40(5250) \\ 60(\frac{11,000}{3}) + 40(2500) \end{array}$ $\begin{array}{ccc}= \\ = \\ =\end{array}$ $\begin{array}{ccc}270,000\\21,000\\320,000\end{arr ay}$

For maximum profit: . $\boxed{x\,=\,\frac{11,000}{3},\;y\,=\,2500}$

. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(d) The maximum profit is: . $\boxed{\$320,000}$
October 30th 2006, 09:41 PM
classicstrings

We did Lin Prog this semester, and it is fairly easy once you get the hang. Those transport problems are a tad more harder. Sometimes when you are writing the constraints you can get mixed up.
October 30th 2006, 11:42 PM
rom

thank you!

thank you very much! thats a great help... i just needed a bit of guidence and you did that perfectly!