Question

1. The total cost consequences of applying insulation to a pipe are cost = 32 t + 2/t where t is insulation thickness. Compute the value of t that minimizes cost.

Answer 1

Given,

TC = 32t + 2/t

Incase of minimization, 1^st order derivative of TC would be 0 and 2^nd order derivative must be greater than 0.

TC = 32t + 2/t

1^st order Derivative of TC = (d/dt) (32t + 2/t)

0 = 32 – 2/t^2

Now, by rearranging

2/t^2 = 32

32t^2 = 2

t^2 = 2/32

t^2 = 0.0625

t = square root 0.0625 = 0.25

now,

2^nd order derivative of TC would be the derivative of 1^st order derivative, which is (32 – 2/t^2).

Derivative of (32 – 2/t^2) is (0 + 4/t^3 =) 4/t^3 is positive, means giving the value greater than 0.

Therefore, this is the case of minimization surely.

Answer: required value of t is 0.25.