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ikadub [295]
3 years ago
9

Suppose y varies directly with x,and y=30 when x=3.what direct variation equation relates x and y.

Mathematics
1 answer:
Olenka [21]3 years ago
8 0
Hello here is a solution :

<span>direct variation equation relates x and y is : y = 10x</span>
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If C(x) is the cost of producing x units of a commodity, then the average cost per unit is c(x) = C(x)/x. Consider the cost func
nadya68 [22]

Answer:

a) C(1,000)=288,491.11 $

b) c(1,000)=288.49 $/u

c) dC/dx(1,000)=349.74 $/u

d) x=100 u

e) c=220 $/u

Step-by-step explanation:

(a) Find the total cost at a production level of 1000 units.

C(x) = 2,000 + 160x + 4x^{3/2}\\\\C(1,000)=2,000 + 160(1,000) + 4(1,000)^{3/2}\\\\C(1,000)=2,000+160,000+126,491.11= 288,491.11

(b) Find the average cost at a production level of 1000 units.

c(x)=C(x)/x\\\\c(1,000)=C(1,000)/1,000=288,491.11/1,000=288.49

(c) Find the marginal cost at a production level of 1000 units.

\frac{dC}{dx} =0+160+4*(3/2)x^{3/2-1}=160+6x^{1/2}\\\\dC/dx|_{1,000}=160+6*(1,000)^{1/2}=160+189.74=349.74

(d) Find the production level that will minimize the average cost.

c=2,000x^{-1}+160+4x^{1/2}\\\\dc/dx=2,000(-1)x^{-2}+0+4(1/2)x^{-1/2}=0\\\\-2,000x^{-2}+2x^{-1/2}=0\\\\2x^{-1/2}=2,000x^{-2}\\\\x^{-1/2+2}=1,000\\\\x^{3/2}=1,000\\\\x=1,000^{2/3}=100

(e) What is the minimum average cost?

c(x) = 2,000/x + 160 + 4x^{1/2}\\\\C(100)=2,000/100 + 160 + 4(100)^{1/2}=20+160+40=220

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The increasing intervals are listed as (-1, 0) u (1, infinity) not because -1 and 1 are not included points of the function, but because the function is not differentiable in those points, so you can't tell whether the function is increasing or decreasing there.

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