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Minimizing cost. A company uses the formula C9x)=0.02x^2-3.4x+150 to model the unit cost in dollars for producing x stabilizer bars. For what number of bars is the unit cost at its maximum? AND what is the unit cost at that level of production?

2006-10-05 11:41:50 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

C(x)=0.02x^2-3.4x+150

In the part where you said "unit cost at its maximum" I think you mean minimum since it is an upward parabola. It wouldn't have a single maximum, plus you also mentioned minimizing cost earlier. So I'll assume you mean minimum.

Take the derivative (a function representing the slope). For this part all you have to remember is the derivative of ax^n is nax^(n-1). And constants disappear.

So in your case the derivative is:
C'(x) = 2(0.02)x - 3.4

Set this to zero (since the minimum occurs where the slope is zero):
0.04x - 3.4 = 0

Solve for x:
0.04x = 3.4
x = 340/4
x = 85

That's your minimum point. Now plug x = 85 into the original equation to get the cost.

C(85) = 0.02(85)^2 -3.4(85) +150
C(85) = 144.50 - 289 + 150
C(85) = 294.50 - 289
C(85) = $5.50 per bar (if these are dollars)

Alternatively, you can graph the equation and realize that the minimum point is at x = 85 and figure the cost the same way, but derivatives are much easier.

2006-10-05 11:46:44 · answer #1 · answered by Puzzling 7 · 0 0

1) Determine slope of cost and set it to zero
dC/dx = 0.04x - 3.4 = 0

2) Solve for x
x = 85

3) Plug back into cost function
C(85) = 0.02(85^2) - 3.4(85) + 150 = 5.5

2006-10-05 18:45:07 · answer #2 · answered by lufen 3 · 0 0

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