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lamthornhill
Group Title
if the cost, C, for manufacturing x units of a certain product is given by C=x^25x+70, find the number of units manufactured at a cost of $8620
 2 years ago
 2 years ago
lamthornhill Group Title
if the cost, C, for manufacturing x units of a certain product is given by C=x^25x+70, find the number of units manufactured at a cost of $8620
 2 years ago
 2 years ago

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richyw Group TitleBest ResponseYou've already chosen the best response.1
well \(C(x)=x^25x+70\) and you know the cost is $8620 so \[x^25x+70=8620\]now solve for x. Do you know how to do that?
 2 years ago

lamthornhill Group TitleBest ResponseYou've already chosen the best response.0
no, that is where I'm having the problem cause I don't understand algebra
 2 years ago

Sheng Group TitleBest ResponseYou've already chosen the best response.0
dw:1352089616943:dw
 2 years ago

Sheng Group TitleBest ResponseYou've already chosen the best response.0
quadratic formula, where a = 1, b= 5, c = 8550
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
alright well the first thing you usually want to do is write this in general form. To do this you subtract 8620 from both sides
 2 years ago

lamthornhill Group TitleBest ResponseYou've already chosen the best response.0
I have no idea what you are talking about
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
\[x^25x+708620=0\]\[x^25x8850=0\]Now you want to find the roots to that. First you want to see if it can be factored, if it can then that's great and if not you need to use the quadratic formuala.
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
i'm getting there!
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
the quadratic formula finds the roots of quadratic polynomials. For a quadratic polynomial in general form, that is \(f(x)=ax^2+bx+c=0\), the quadratic formulas is \[\frac{b\pm\sqrt{b^24ac}}{2a}\]
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
so looking at the equation we have \(x^25x8850\), you can pick out the parameters a,b and c.
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
here the \(x^2\) term has nothing in front of it, so \(a=1\), the \(x\) term has a 5 in front of it so \(b=5\), and \(c=8850\). Does that make sense?
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
sorry \(c=8850\)
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
now plug those numbers into the the quadratic equation.
 2 years ago

lamthornhill Group TitleBest ResponseYou've already chosen the best response.0
sorry I'm not responding....... i'm writing and studying this so I can understand it
 2 years ago

lamthornhill Group TitleBest ResponseYou've already chosen the best response.0
I don't even know what the quadratic equation is
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
yup MEMORIZE the quadratic formula! also do you understand what the \(\pm\) sign means?
 2 years ago

lamthornhill Group TitleBest ResponseYou've already chosen the best response.0
no....... this is all new to me....... I didn't take algebra in high school
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
alright well a quadratic EQUATION is a polynomial equation of the second degree. The initial problem you have is a quadratic equation. The quadratic FORMULA is the formula I gave you. You use the quadratic formula to find the roots of quadratic equations.
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
This is the quadratic FORMULA.\[\frac{b\pm\sqrt{b^24ac}}{2a}\]
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
the \(\pm\)is just used to save space, it means that this is actually two formulas...haha\[\frac{b+\sqrt{b^24ac}}{2a}\]\[\frac{b\sqrt{b^24ac}}{2a}\]
 2 years ago

lamthornhill Group TitleBest ResponseYou've already chosen the best response.0
I have been on this type of problem now for 2 hours
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
but this equation is only useful if your equation is in general form. That was the first step that we did (when we subtracted 8620)
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
right so now just plug the numbers in \[\frac{b\pm\sqrt{b^24ac}}{2a}=\frac{(5)\pm\sqrt{(5)^24(1)(8550)}}{2(1)}\]
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
this gives \[x=\frac{5+\sqrt{25+4(8550)}}{2}\]\[x=\frac{5\sqrt{25+4(8550)}}{2}\]
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
For the context of the question though you can't have negative sales, you you only need to plug in the top one...
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
sorry! how could I have forgotten this, if you were confused before, the quadratic equation gives you x... I'll put in an x= there for you :D. Anyways I don't have a calculator, so plug those two above in.\[x=\frac{b\pm\sqrt{b^24ac}}{2a}\]
 2 years ago

lamthornhill Group TitleBest ResponseYou've already chosen the best response.0
for the top I came up 34225
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
nope that's not correct, you must have messed up on the calculator. Hand on i'll just check wolfram...
 2 years ago

lamthornhill Group TitleBest ResponseYou've already chosen the best response.0
ok... I'm not good with math.....
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
well we already finished the math, now it's just using the calculator :). But the two roots of this polynomial are x=90 and x=95. Obviously the answer to the question is x=95, because you can't manufacture a negative number of products. Just you know you can always verify this on wolfram alpha http://www.wolframalpha.com/input/?i=x%5E25x%2B70%3D8620
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
It might seem compplicated but really all you did was two steps 1)Put equation in general form 2)Find the roots (in this case we found the roots using the quadratic equation).
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
No worries, and good luck with your math studies! I only learned algebra after high school as well. It just takes a LOT of work! It is sure worth it though.
 2 years ago
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