SOLUTION: Question: Write a cost function for the problem. Assume that the relationship is linear. Marginal cost, $40; 70 items cost $3800 to produce a) C(x)=14x+3800 b) C(x)=40x+1000

Algebra ->  Functions -> SOLUTION: Question: Write a cost function for the problem. Assume that the relationship is linear. Marginal cost, $40; 70 items cost $3800 to produce a) C(x)=14x+3800 b) C(x)=40x+1000       Log On


   



Question 164277: Question: Write a cost function for the problem. Assume that the relationship is linear.
Marginal cost, $40; 70 items cost $3800 to produce
a) C(x)=14x+3800
b) C(x)=40x+1000
c) C(x)=40x+3800
d) C(x)=14x+100

Question: Assume that the sales of a certain appliance dealer are approximated by a linear function. Suppose that sales were $10,500 in 1982 and $67,000 in 1987. Let x=0 represent 1982. Find the equation giving yearly sales S(x).
a) S(x)=11,300x+67,000
b) S(x)=56,500x+10,500
c) S(x)=56,500x+67,000
d) S(x)=11,300x+10,500

Mike

Answer by watchmath(2) About Me  (Show Source):
You can put this solution on YOUR website!
Marginal cost means additional cost for additional item which is $40/item. But this is exactly the slope. You have another information that (70, 3800) is on the line.
So by the point slope formula
y-y_1=m(x-x_1)
we have
C-3800= 40(x-70)
C(x)=40(x-70)+3800
C(x)=40x +1000
So the answer ============> (b)
The given two data about the sales in 1982 and 1987 tell us that the point
(1982, 10,500) and (1987, 67,000) are on the line.
The slope then equals (67,000 - 10,500)/(1987 - 1982) =56,500/5=11300.
We have
S(x)=11,300 x+b
since x=0 correspond to 1982 and we now that S=10,500 in 1982 then
10,500 =11,300 (0)+b ========> b=10,500 and therefore
S(x)=11,300x+ 10,500.
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