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11111nata11111 [884]
3 years ago
9

The cost C, in dollars, of building m sewing machines at Sienna’s Sewing

Mathematics
2 answers:
Elenna [48]3 years ago
7 0

OK.  So the cost to manufacture any number 'm' machines is

<span>                               C(m) = 20m^2 - 830m + 15,000 .

Whatever number of machines you're interested in, you write
that number in place of 'm', and this equation tells you the cost
for that many.

Examples:

-- The cost to manufacture zero sewing machines ... what the
company had to invest in equipment and building space before
they could even start manufacturing anything:

                    </span><span><span>C(m) = 20m^2 - 830m + 15,000

                    C(0)  =  20(0)²  -  830(0)  +  15,000  =  15,000 .

</span>-- The cost to manufacture one sewing machine ... buy the
building, set up the manufacturing equipment, and turn out
the first one:

                      </span><span>C(m) = 20m^2 - 830m + 15,000

                      C(1)  =  20(1)²  -  830(1)  +  15,000  =  14,190 .

Now, part-a) wants to know the cost to build 75 sewing machines. 
If you've been paying attention so far, you know you have to take
the same equation, and write '75' in place of 'm'.

                       </span><span><span>C(m)  =  20m^2  -  830m  +  15,000

                       C(75)  =  20(75)²  -  830(75)  +  15,000</span>

                                   = 20(5,625) - 830(75) + 15,000

                                   = 112,500  -  62,250  +  15,000  =  65,250 .
===================

Now you need to find the number of sewing machines
that can be built for the lowest total cost.

</span><span>I'm sure you noticed that the equation for the cost  C(m)  is a
quadratic equation.  So if you drew it on a graph, it would be
a parabola.  It would have a minimum value at some 'm', and
for greater 'm', it would start going up again.
 
(Why should your cost start increasing past some number of
sewing machines ?  Well, maybe the manufacturing equipment
is starting to wear out, and needs repair more often.</span>  All of that
is actually built into the equation for C(m) . )

Now, I'm not sure what method you've learned for finding the
minimum value of a parabola (quadratic equation).  Here are
the two ways I know:

Way #1).  If you've had some pre-calculus, then you'll take the
derivative of the equation, set the derivative equal to zero, and
that leads you to the minimum:

The equation:              <span>  C(m) = 20m^2 - 830m + 15,000

Its first derivative:          C'(m) = 40m - 830

'C'; is minimum when C'=0 :      40m - 830 = 0

Add 830 to each side:                40m          = 830

Divide each side by  40 :                m          = 20.75

The number of sewing machines manufactured for the
minimum total cost is  20  or  21 .
</span>
Way #2).  Really the same as Way-#1 but it's not called 'derivative'.

I looked online for rules of parabolas, and found the one that
you may have learned to use:

       For the quadratic expression    Ax² + Bx + C ,
       the axis (midline) of the parabola is at
                                                                           x = - B / 2A .

That's exactly what we need.
Our equation is                            <span>C(m) = 20m^2 - 830m + 15,000

so the axis of the parabola is at        m  =  - (-830)/2(20)

                                                                   =      830/40  =  20.75 .

Same as Way-1 .
</span>
bulgar [2K]3 years ago
6 0
So basically function of m (f(m) or in this case C(m)) means the price
so just input the value you put for m for all the other m's in the problem
ex. if you had f(x)=3x and you wanted to find f(4) then you replace and do f(3)=3(4)=12 so f(3)=12 and so on



A. cost of 75 sewing machines
75 is the number you replace m with
C(75)=20(75)^2-830(75)+15,000
simplify
20(5625)-62250+15000
112500-47250
65250
the cost for 75 sewing machines is $65,250


B. we notice that in the equation, that the only negative is -830m
so we want anumber that will be big enough to make -830m destroy as much of the other posities a possible

-830m+20m^2+15000
try to get a number that when multiplied by 830, is almost the same amount as or slightly smaller than 20m2+15000 so we do this
830m<u><</u>20m^2+15000
subtract 830m from both sides
0<u><</u>20m^2-830m+15000
factor using the quadratic equation which is
(-b+ the square root of (b^2-4ac))/(2a) or (-b- the square root of (b^2-4ac))/(2a)
in 0=ax^2+bx+c so subsitute 20 for a and -830 for b and 15000 for c
you will get a non-real result I give up on this meathod since it gives some non real numbers so just guess

after guessing and subsituting, I found that the optimal number was 21 sewing machines at a cost of 6420
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