#### To determine

**(a)**

**To explain:**

Why ∫0tfsds represents the loss in the value of the machine over the period of time t since the last overhaul.

#### Answer

F't=ft represents the rate of depreciation, so Ft represents the loss inthe value on the interval 0, t.

#### Explanation

**1) Concept:**

The Fundamental Theorem of Calculus: Suppose f is continuous on [a, b], then

∫abfxdx=Fb-F(a), where F is antiderivative of f, that is F'=f.

**2) Given:**

ft=rate of depression

t=time in months

**4) Calculations:**

Let Ft=∫0tfsds.

Then by Fundamental Theorem of Calculus,

F't=f(t)

and ft=rate of depreciation

So F(t) represents the loss in value on the interval 0, t.

**Conclusion:**

F't=ft represents the rate of depreciation, so Ft represents the loss in the value on the interval 0, t.

#### To determine

**(b)**

**To find:**

What does C representand why would the company want to minimize C?

#### Answer

C(t) represents the average amount of expenditure per unit of time t on the interval 0, t.

The company wants to minimize the average expenditure because they want to spend as little money as possible.

#### Explanation

**1) Given:**

t=time in months

**2) Calculations:**

Let C=Ct be given by

Ct=1tA+∫0tfsds

But ∫0tfsds=F(t)

So,Ct=A+F(t)t

Since F(t) represents the loss in the value on the interval [0, t], and t represents time, Ct represents the average amount of expenditure per unit time t on the interval [0, t].

The company wants to minimize the average expenditure because they want to spend as little money as possible.

**Conclusion:**

C(t) represents the average amount of expenditure per unit of time t on the interval 0, t.

The company wants to minimize the average expenditure because they want to spend money as little as possible.

#### To determine

**(c)**

**To show:**

C has minimum value at the numbers t=T, where, CT=fT.

#### Answer

C has minimum value at the numbers t=T, where CT=fT.

#### Explanation

**1) Given:**

Ct=1tA+∫0tfsds

**2) Calculations:**

Consider,

Ct=1tA+∫0tfsds

To find minimum, take the derivative of the above equation and set it equal to zero.

C't=ddt [1t( A+∫0tfsds )]

Use the product rule to find the derivative

ddt1t(A+∫0tfsds)+ddt( A+∫0tfsds)·1t

By taking derivative,

-1t2·(A+∫0tfsds)+0+ft·1t

Factoring out 1t from the above equation

1t[-1t(A+∫0tfsds)+(ft)]=C'(t)

Now, since to find minimum, set this equation of C'(t) =0;

1t[-1t(A+∫0tfsds)+(ft)]= 0

After multiplying both sides by t,

-1t(A+∫0tfsds)+(ft)= 0 If any time passes at all, then t is not equal to zero and 1t is defined for all t.

Leave f(t) and subtract everything to the other side

ft= --1t(A+∫0tfsds)

After simplifying,

ft= 1t(A+∫0tfsds)

This is the same equation for C(t) that was originally given so

ft= 1t(A+∫0tfsds)= C(t)

C't=0 Only at a specific time t.

Call, t = T which gives

CT=f(T)

**Conclusion:**

C has minimum value at the numbers t = T, where CT=fT.