# maximize profit word problems, yikes

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• Mar 24th 2009, 08:39 AM
as488744
maximize profit word problems, yikes
Everyone's favorite type of math question, word problems (Clapping)

After working on these for a few hours and getting nowhere! I even tried having my brother help me (Giggle). I would appreciate any and all help with the following word problems that will be the death of me ((Shake)):

1. (Buying and selling beef.) A steer weighing 300 pounds gains 8 pounds per day and costs 1.00 dollar a day to keep. You bought this steer today at a market price of 1.50 dollars per pound. However, the market price is falling 2 cents per pound per day. When should you sell the steer in order to maximize your proﬁt?

2. (Maximizing production.) An orange grower in California hires migrant workers to pick oranges during the season. He has 12 employees, and each can pick 400 oranges per hour. He has discovered that if he adds more workers, the production per worker decreases due to lack of supervision. When x new workers (above the 12) are hired, each worker picks
400 − 2x 2 oranges per hour.

(a) Write the total number N (x) of oranges picked per hour as a function of x;
(b) Find the domain of N (x);
(c) Graph the function N (x);
(d) Estimate the maximum number of oranges that can be picked per hour;
(e) Find the number of employees involved in part (d).

3. (Biology.) The volume V of a lung is directly proportional to its internal surface area A. A lung with volume 400 cm3 from a certain species has an average internal surface area of 100 cm2 . Find the volume of a lung of a member of this species if the lung’s internal surface area is 120 cm2 .

Thank you math gurus out there! (Rock)
• Mar 24th 2009, 01:20 PM
Opalg
Quote:

Originally Posted by as488744
Everyone's favorite type of math question, word problems (Clapping)

Yay, my favourite too! (Clapping)

Quote:

Originally Posted by as488744
1. (Buying and selling beef.) A steer weighing 300 pounds gains 8 pounds per day and costs 1.00 dollar a day to keep. You bought this steer today at a market price of 1.50 dollars per pound. However, the market price is falling 2 cents per pound per day. When should you sell the steer in order to maximize your proﬁt?

Suppose you sell the steer after x days.

How much will it weigh by then? (300 + 8x pounds)

What will the price per pound be by then? (In cents, 150 – 2x ¢)

How much will you get for selling it? (Multiply the number of pounds by the price per pound.)

How much will you have had to pay for the steer plus the cost of keeping it? (In cents, 300 times 150 for the original cost, plus 100x for the cost of keeping it.)

So how much will the profit be? (Income from sale, less total costs.)

Finally, you have a formula for the profit (in terms of x). Now the problem has become purely mathematical, and you just have to figure out what its maximum value is.
• Mar 24th 2009, 03:23 PM
as488744
Still not sure...
Here is what I did:

P(x)= (300+8x)(150-2x) - ((300*150)+100x)
0 = (45000 + 600x - 16x^2) - 45000 + 100x
0 = -16x^2 + 100x
0 = x(16x - 100)
6.25 = x

However, when I plugged 6.25 in for x it doesn't give me 0 for P(x) (Headbang)

Am I missing a step here or something?
• Mar 24th 2009, 03:59 PM
rtblue
Hello there. Well. I started at number 3 (thats just how cool i am :D)

ok. The key words in this is: Directly Proportional.

This is VERY important. The formula for direct variation is: Y=KX

Where K is the constant multiplier

Where Y is the dependent variable

Where X is the independent variable

Now. The first objective is to find the constant multiplier.

Apparently the volume is the dependent variable, because at the end of the problem, the question is asking for the volume. Lets plug this in.

400(Volume)=100(Surface area)K

400=100K

From solving, we can determine that K=4

Now, plug this back into the problem, where it says that the internal surface area is 120 cm.

Y=4(120)

Y=480

So the volume of the lung is 480cm3 when the internal surface area is 120 cm.

YAY :D
• Mar 25th 2009, 07:56 AM
Opalg
Quote:

Originally Posted by as488744
Here is what I did:

P(x)= (300+8x)(150-2x) - ((300*150)+100x)
P(x) = (45000 + 600x - 16x^2) - 45000 100x Should be –100x, not +100x; and it's P(x), not 0 (P(x) is the profit, and you don't want that to be 0!).
0 = -16x^2 + 100x This line should say P(x) = 500x – 16x^2
0 = x(16x - 100)
6.25 = x

However, when I plugged 6.25 in for x it doesn't give me 0 for P(x) (Headbang)

Am I missing a step here or something?

So you want to find the value of x that maximises P(x) = 500x – 16x^2. You do this either by graphing P(x) or (if this is a calculus course) by differentiating it and putting the derivative equal to 0.