For the given cost function Cx 259 squareroot x x2125000 f

For the given cost function C(x) = 259 squareroot x + x^2/125000 find The cost at the production level 1800 The average cost at the production level 1800 The marginal cost at the production level 1800 The production level that well minimize the average cost. The minimal average cost.

Solution

given cost C(x)=250x +(x²_/125000)

a)x =1800

C(1800)=2501800 +(1800²_/125000)

C(1800)=10632.52

cost at production level 1800 is 10632.52

b)cost C(x)=250x +(x²_/125000)

average cost AC(x)=(C(x))_/x

AC(x)=(250x +(x²_/125000))_/x

AC(x)=(250_/x) +(x_/125000)

x=1800

AC(1800)=(250_/1800) +(1800_/125000)

AC(1800)=5.907

average cost ot production level 1800 is 5.907

c)cost C(x)=250x +(x²_/125000)

marginal cost C\'(x)=(250_/2x) +(2x_/125000)

marginal cost C\'(x)=(125_/x) +(x_/62500)

x =1800

marginal cost C\'(1800)=(125_/1800) +(1800_/62500)

marginal cost C\'(1800)=2.975

d)average cost AC(x)=(250_/x) +(x_/125000)

dAC\'(x) =(-125_/x^3/2) +(1_/125000)

for minimum average cost dAC\'(x) =0

(-125_/x^3/2) +(1_/125000)=0

(125_/x^3/2) =(1_/125000)

x^3/2=125*125000

x=62500

production level that will minimise average cost =62500

e) minimal average cost =AC(62500)

=(250_/62500) +(62500_/125000)

=1+0.5

=1.5

minimal average cost =1.5