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sveta [45]
3 years ago
8

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

of 100 units can be sold per month.
How many units should be sold to maximize monthly profit? units

What is the maximum monthly profit when this number of units are sold? (Round your answer to the nearest whole number.) dollars

(Hint: When you solve for the critical values in this question, you'll have to solve a quadratic equation. You can do that using the quadratic formula, or you can do it by factoring.)

How to get this answer?
Mathematics
1 answer:
wlad13 [49]3 years ago
6 0

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information:P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000, where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2

Equating it to zero we get,

x^2 + 36x - 6400 = 0

We use the quadratic formula to find the values of x:

x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}, where a, b and c are coefficients of x^2, x^1 , x^0 respectively.

Putting these value we get x = -100, 64

Now, again differentiating

\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x

At x = 64,  \displaystyle\frac{d^2(P(x))}{dx^2} < 0

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

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A foreman for an injection-molding firm admits that on 55% of his shifts, he forgets to shut off the injection machine on his li
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2 years ago
PLEASE HELP! I HAVE TWO QUESTIONS AND WILL GIVE 100PTS AND BRAINLIEST IF CORRECT!
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Answer:

1) $283.50

2) 65.21 meters.

Step-by-step explanation:

Part 1)

We know that a spool of ribbon holds 6.75 meters.

A craft club buys 21 spools.

So, the total length of ribbon bought is:

T=21(6.75)=141.75\text{ meters}

Each meter costs $2. So, the total cost will be:

141.75(2)=\$ 283.50

Part 2)

A spool holds 6.75 meters.

We bought 21 of them, so that total length we have is 141.75 meters.

The club used 76.54 meters.

So, the amount of ribbon that is left is represented by:

141.75-76.54

Subtract:

=65.21\text{ meters}

So, 65.21 meters of ribbon will be left.

7 0
3 years ago
Read 2 more answers
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