The marginal cost of a product is modeled by

dC/dx= 16/cube root 16x + 3
where x is the number of units. When x = 17, C = 120.
(a) Find the cost function. (Round your constant term to two decimal places.)
c=

(b) Find the cost of producing 90 units. (Round your answer to two decimal places.)
$

To find the cost function, let's start by integrating the given marginal cost function:

∫dC/dx dx = ∫(16/(16x)^(1/3) + 3) dx

Using the power rule of integration, we have:

C = ∫16/(16x)^(1/3) dx + ∫3 dx

Integrating both terms separately:

C = 3∫ dx + 16∫(16x)^(-1/3) dx

Simplifying:

C = 3x + 16(3/2)((16x)^(2/3)) + K

Where K is the constant of integration.

Since we are given that C = 120 when x = 17, we can substitute these values into the equation to solve for K:

120 = 3(17) + 16(3/2)((16(17))^(2/3)) + K

120 = 51 + 16(3/2)((16^2)^(2/3)) + K

120 = 51 + 16(3/2)(4^2) + K

120 = 51 + 16(3/2)(16) + K

120 = 51 + 24(16) + K

120 = 51 + 384 + K

120 = 435 + K

K = 120 - 435

K = -315

Therefore, the cost function is:

C = 3x + 16(3/2)((16x)^(2/3)) - 315

(a) The cost function is c = 3x + 24(2x)^(2/3) - 315.

Now, to find the cost of producing 90 units, we plug in x = 90 into the cost function:

c = 3(90) + 24(2(90))^(2/3) - 315

c = 270 + 24(180)^(2/3) - 315

c ≈ 270 + 24*(167.9779) - 315

c ≈ 270 + 4031.4696 - 315

c ≈ 3986.4696

(b) The cost of producing 90 units is approximately $3986.47.

To find the cost function, we will integrate the given marginal cost function with respect to x.

(a)
Given: dC/dx = 16/(16x)^(1/3) + 3

To find C, integrate both sides of the equation with respect to x:
∫dC = ∫(16/(16x)^(1/3) + 3) dx

Integrating the first term on the right side requires applying the power rule of integration in reverse:
∫(16/(16x)^(1/3)) dx = 16 ∫(1/(16x)^(1/3)) dx = 16 ∫(x^(-1/3))/16 dx = ∫x^(-1/3) dx = 3x^(2/3) + C1.

The second term on the right side is a constant multiple, so it can be integrated directly:
∫3 dx = 3x + C2.

Now we can combine both results:
C = C(x) = 3x^(2/3) + 3x + C1 + C2.

Given the condition x = 17, C = 120, we can substitute these values and solve for the constant terms:
120 = 3(17)^(2/3) + 3(17) + C1 + C2.

Simplifying the equation:
120 = 3(4913^(1/3)) + 51 + C1 + C2,
120 = 3(17) + 51 + C1 + C2,
120 = 102 + C1 + C2,
18 = C1 + C2.

To round the constant terms to two decimal places, we can let C1 = 9 and C2 = 9:
C = 3x^(2/3) + 3x + 18.

Therefore, the cost function is C = 3x^(2/3) + 3x + 18.

(b)
To find the cost of producing 90 units, substitute x = 90 into the cost function:
C(90) = 3(90)^(2/3) + 3(90) + 18.

Calculating the expression:
C(90) = 3(138.63) + 3(90) + 18,
C(90) ≈ 415.89 + 270 + 18,
C(90) ≈ 703.89.

Therefore, the cost of producing 90 units is approximately $703.89.

I read that as

dC/dx = 16/(16x+3)^(1/3)

then C = (3/2)(16x + 3)^(2/3 + c
when x = 17, C=120
120 = (3/2)(275)^(2/3) + c

c = ....

Once you have c, you know C = ....
sub in x = 90