If the cost, c, for manufacturing x units of a certain product is given by C=x^2-15x+55, find the number of unites manufactured at a cost of $8555
please help... please?
2007-03-11 14:35:26 · 6 answers · asked by xrandomnessx 2 in Science & Mathematics ➔ Mathematics
This is a simple quadratic equation.
Firstly plug in the values:
8555 = x^2-15x+55
simplifying, x^2-15x-8500 = 0
factorising
(x-100) (x+85) = 0
thus x= 100 or - 85. Since it cant be -85, the answer is 100.
100 units are manufactured at cost of $8555
2007-03-11 15:31:00 · answer #1 · answered by Ratbag 2 · 0⤊ 0⤋
So 8555 = x^2 - 15x + 55.
To solve this you must make it equal to zero. Do this by subtracting 8555 from both sides.
0 = x^2 - 15x - 8500
Then factor it by FOIL: 0 = (x - 100)(x + 85)
Use this to find x, remembering that x cannot be negative.
2007-03-11 21:41:11 · answer #2 · answered by hayharbr 7 · 0⤊ 0⤋
8555 = x^2 -15x +55
x^2 -15x -8500 = 0
(x-100)(x+85) = 0
x= 100 or -85
since we know there can't be any negative number for real life problems, the answer is 100
2007-03-11 21:39:37 · answer #3 · answered by inane person 2 · 0⤊ 0⤋
8555=x^2-15x+55
0=x^2-15x-8500
use quadratic formula to get
100
or factor -8500 into (100)*(-85)
now realize that 100-85=15 so,
(x+85)(x-100)=0
x= -85 or 100
2007-03-11 21:41:13 · answer #4 · answered by lcjjr87 2 · 0⤊ 0⤋
C = x^2-15x+55,
8555 = x^2-15x+55
Therefore,
x^2-15x - 8500 = 0
Use Quadratic formula.
Solve for X.
Check this site for help...http://mathworld.wolfram.com/QuadraticEquation.html
2007-03-11 21:53:06 · answer #5 · answered by carrier_anomaly_detected 2 · 0⤊ 0⤋
solve for x the 2nd degree equation
x^2-15x-8500=0 (you should know the formula)
2007-03-11 21:42:28 · answer #6 · answered by santmann2002 7 · 0⤊ 0⤋