anonymous
  • anonymous
please help! I need to pass algebra to graduate! widget wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. the company also discovered that it cost $16 to produce 4 widgets, ans $48 to produce 10 widgets. find the total cost of producing 8 widgets.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
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Curry
  • Curry
So 16 = a(4)^2 - b4 -c and 48 = a(10)^2 -b(10) - c
Curry
  • Curry
Correction, those minus signs should be positive. so solve for those two equations, and the Cs cancel immediately. so your'e left with 16 =16a+4b. and 48 = 100a+10b.
Curry
  • Curry
you can figure out a and b frmo there, and then you can find c once you have a and b. Lastly, when you have the three constants, (a,b,c), you can just plug in 8 for x into the quadratic eqauation. let me know if you need more help.

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anonymous
  • anonymous
yeah Im so lost I don't even know what this means :(
anonymous
  • anonymous
@Curry
anonymous
  • anonymous
do you still need help?
anonymous
  • anonymous
yes!
anonymous
  • anonymous
ok so can you tell me what the general form of a quadratic function is?
anonymous
  • anonymous
if not thats ok, i need to see what you know and what you dont
anonymous
  • anonymous
I don't
anonymous
  • anonymous
ok, the general form of a quadratic function is this ax^2 + bx + c a = some number that is not 0 b = some number c = some number for example 2x^2 + 5x + 4
anonymous
  • anonymous
|dw:1433582094129:dw|
anonymous
  • anonymous
those two are equivalent, the one on the left is a way of writing it that we can easily type with a keyboard
anonymous
  • anonymous
so like a= 1/2 b= -2 c= 18 this is what I got so far
anonymous
  • anonymous
ok so if a = 1/2 b = -2 and c = 18 then what would the general form look like?
anonymous
  • anonymous
1/2x^2+(-2x)+18?
anonymous
  • anonymous
perfect
anonymous
  • anonymous
and simplying you get 1/2x^2 - 2x + 18
anonymous
  • anonymous
ok good i think you are ready to set up the problem
anonymous
  • anonymous
ok so now it says the cost, c(x), of making x widgets is quadratic, so lets write it out c(x) = ax^2 + bx + c x is the input of the function, c(x) is the output of the function
anonymous
  • anonymous
does this make sense?
anonymous
  • anonymous
so you will input the number of widgets, x, into the function and the output will be the cost, c(x)
anonymous
  • anonymous
does all of this make sense so far?
anonymous
  • anonymous
can you set it up for me this is a little confusing
anonymous
  • anonymous
you should set it up, so that when you are taking a test, you can do it yourself
anonymous
  • anonymous
cost = c(x) = output number of widgets = x = input
anonymous
  • anonymous
cost is the same thing as c(x) number of widgets is the same thing as x
anonymous
  • anonymous
ok so lets start with the first example
anonymous
  • anonymous
they said the cost was 16, so can you plug the 16 into this equation? c(x) = ax^2 + bx + c
anonymous
  • anonymous
so c(x) = 16x^2 -2x +18
anonymous
  • anonymous
no remember cost is the same thing as c(x)
anonymous
  • anonymous
if cost is 16 and cost is c(x) what is c(x)?
anonymous
  • anonymous
16
anonymous
  • anonymous
perfect, so c(x) = 16 lets plug it in
anonymous
  • anonymous
c(x) = ax^2 + bx + c now im plugging in c(x) = 16 16 = ax^2 + bx + c
anonymous
  • anonymous
ok yes that makes sense
anonymous
  • anonymous
good
anonymous
  • anonymous
so now remember widgets is the same thing as x, and it says it costs 16 to produce 4 widgets
anonymous
  • anonymous
so we have 16 = ax^2 + bx + c can you plug 4 in here in the correct place?
anonymous
  • anonymous
x is the number of widgets the number of widgets is 4 so x must be?
anonymous
  • anonymous
and what about a b and c?
anonymous
  • anonymous
dont worry about those yet
anonymous
  • anonymous
step by step
anonymous
  • anonymous
so whats our equation look like now?
anonymous
  • anonymous
16=a(4)^2 -2(4) +18?
anonymous
  • anonymous
perfect
anonymous
  • anonymous
ok now set up another equatoin in the exact same way, with the other information cost is 48 to produce 10 widgets
anonymous
  • anonymous
oh actually that last one should be 16 = a(4)^2 + b(4) + c sorry
anonymous
  • anonymous
yes I thought so too
anonymous
  • anonymous
can you set up the other equatoin now?
anonymous
  • anonymous
cost is 48 to produce 10 widgets, plug the given values into the equation c(x) = ax^2 + bx + c
anonymous
  • anonymous
48= a(10^2) -b(10) +c
anonymous
  • anonymous
very close, except its +b(10) not -b(10)
anonymous
  • anonymous
sorry I got mixed up because b is -2
anonymous
  • anonymous
ok perfect so we have these two equations: 16 = a(4)^2 + b(4) + c 48 = a(10)^2 + b(10) + c
anonymous
  • anonymous
all good
anonymous
  • anonymous
can you simplify those two equatoins?
anonymous
  • anonymous
I don't know how
anonymous
  • anonymous
ok i will do one you do the other
anonymous
  • anonymous
16 = a(4)^2 + b(4) + c simplifies to this: 16 = a(16) + b(4) + c written in a standard way looks like this: 16 = 16a + 4b + c
anonymous
  • anonymous
alright that was too easy all you did was square the 4
anonymous
  • anonymous
exactly
anonymous
  • anonymous
ok so we have these two equations: 16a + 4b + c = 16 100a + 10b + c = 48
Curry
  • Curry
well actually, hold on.
anonymous
  • anonymous
48= a(10^2) +b10 +c 48= a100 +b10 +c
anonymous
  • anonymous
perfect :)
anonymous
  • anonymous
so now we have the two equations that @Curry posted in the first post
anonymous
  • anonymous
so now we have to find the values of a, b, and c that make those two equations true, do you know how to do that?
anonymous
  • anonymous
nope
Curry
  • Curry
wait, are we missing the initial condition value?
Curry
  • Curry
that's only 2 equations. we have three unknowns.
Curry
  • Curry
hey @billj5, i calculated it ignoring the c too. And i got 36.2 as i posted earlier.
Curry
  • Curry
But strictly speaking, we can't jsut assume that c is 0. if you did, then yes, you can just ignore c. and the answer would be 36.2
anonymous
  • anonymous
ok are you still here sarah?
anonymous
  • anonymous
brb i need to grab something to eat
anonymous
  • anonymous
let me know if you are still here sarah and we can finish the problem
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok great, are you familiar with solving systems of equations?
anonymous
  • anonymous
yes!
anonymous
  • anonymous
since this is a cost function and they didnt give us any information about initial cost before you take into account the number of widgets, lets assume that c = 0 ok?
anonymous
  • anonymous
so we have these two equations: 16a + 4b = 16 100a + 10b = 48 so can you solve that system for a and b?
anonymous
  • anonymous
@Curry you brought up a good point, sorry, I think we can assume c = 0
anonymous
  • anonymous
let me know the values you get for a and b
anonymous
  • anonymous
ok so what do I solve for a?
anonymous
  • anonymous
you solve the system of those 2 equations for a and b
Curry
  • Curry
ye, that's what i did too. Assumed c = 0. This is a bad problem, considering they didn't give us an initial condition. Hence, we're basically forced to omit C. which defeats half the purpose of a quadratic equation.
anonymous
  • anonymous
yes, i agree @Curry
anonymous
  • anonymous
sarah please let me know what you get for the values of a and b
anonymous
  • anonymous
a=-b/4+1 b=-4a+4 now the seconf one
anonymous
  • anonymous
yes what you wrote is correct, however you need to find the number value for both a and b
anonymous
  • anonymous
are you familiar with solving systems of equations by substitution or elimination or both?
anonymous
  • anonymous
hey im running out of time can you tell me the answer and then we can continue working on it so I can learn? and no im not familiar with those
anonymous
  • anonymous
well i think @Curry posted the answer above
anonymous
  • anonymous
i didnt check if it was correct though
anonymous
  • anonymous
can you check and tell me because this test is gonna end and I need to pass. then we can continue learning
anonymous
  • anonymous
what did he say was the answer?
anonymous
  • anonymous
it was wrong... I failed.. and its 3 am im tired im just gonna cut and go to sleep ;c
anonymous
  • anonymous
fill in
Curry
  • Curry
problem with rounding probably...
Curry
  • Curry
unless they assumed a value for c or something.
math&ing001
  • math&ing001
@Curry @billj5 Here the c is the cost we're looking for. A quadratic equation looks like this ax^2+bx+c=0 hence c=ax^2+bx Since c is a function of x, we can write it like this c(x)=ax^2+bx. The rest is pretty much what you said: 16a + 4b = 16 100a + 10b = 48 When solved you get a=2/15 and b=54/15 Calculating the cost for 8 widgets: c(8) = 2/15 * 8^2 + 54/15 * 8 = $37.33
anonymous
  • anonymous
@math&ing001, b is 52/15 and c is not the cost we are looking for, c(x) is the cost we are looking for, c is a constant in the quadratic function which is not what we are looking for, also, if you did what you said it would be -c=ax^2+bx not c=ax^2+bx
math&ing001
  • math&ing001
Here the c is not a constant and is a function of x hence the notation c(x). You're right b=52/15 which makes it $36.27. As for the minus sign, because it's a variable, writing c or -c is absolutely the same.
anonymous
  • anonymous
@math&ing001, no you are confusing c in the general form of a quadratic which is a constant term and c(x) which is the function itself, and c is not the same as -c
math&ing001
  • math&ing001
I'm just trying to teach you something new, but if you're not open to learning then there's not much I can do about it.
anonymous
  • anonymous
@math&ing001, no you are giving wrong information, but thank you
math&ing001
  • math&ing001
Just because you don't understand it, doesn't make it wrong. Anyways, I'm not going to keep this conversation going, it's not taking us anywhere. Have a good day !
anonymous
  • anonymous
@math&ing001, please have any of the mods come in here and verify what you are saying.. i promise that you are wrong, and im very open to learning by the way

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