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masha68 [24]
3 years ago
15

A company sells mattresses for $365each. To produce a batch of nmattresses, there is a cost of $254 per mattress and a fixed or

setup cost of $9,100 for the entire batch. Determine a function that gives the profit in terms of the number of mattresses produced. What is the number of mattresses that must be produced and sold in order for the company to break even
Mathematics
1 answer:
neonofarm [45]3 years ago
8 0
Any function to determine profit is revenue - cost. 

If the number of mattresses made is x, then the revenue made per mattress is 365x. 

Under the same assumption, the cost per mattress is 254x + 9100, adding the setup costs to the cost per mattress.

Thus, the function is f(x) = 365x - (254x + 9100). We can simplify this function.

First, lets expand the negative (the same as multiplying each value by -1).
f(x) = 365x - 254x - 9100

Next, we'll combine like terms.
f(x) = 111x - 9100
Therefore, the function for profit is f(x) = 111x - 9100.

Continuing, we need to determine how many mattresses the company needs to break even. Breaking even means there is no profit, but also no loss. Otherwise, they make 0 total. So, we can set the function we just found to 0 and solve for x.


Set the function to 0.
0 = 111x - 9100

Add 9100 to both sides.
9100 = 111x

Divide both sides by 111.
81.98 = x

Rounding this number up, it would take 82 mattresses to break even.
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What is the coefficient of x2y3 in the expansion of (2x + y)5?
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Option C:

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Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

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