#### To determine

**(a)**

**To show,**

The critical numbers of C occur at the numbers t where Ct=ft+gt.

#### Answer

The critical numbers of C occur at the numbers t where Ct=ft+gt.

#### Explanation

**1) Concept:**

Use The Fundamental Theorem of Calculus to find the critical point.

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) Calculations:**

Consider,

Ct=1t∫0tfs+gs ds

To find the critical point, differentiate the above equation with respect to t using the power rule

C't= -1t2∫0tfs+gs ds +1t{ddt(∫0tfs+gsds} .

Using Fundamental Theorem of Calculus,

C't= -1t2∫0tfs+gs ds +1tft+gt

In order to find the critical point, equate C'(t) to zero

Thus,C't= -1t2∫0tfs+gs ds +1tft+gt=0

1t2∫0tfs+gs ds = 1tft+gt.

Multiplying both sides by t,

1t∫0tfs+gs ds = ft+gt.

But

Ct=1t∫0tfs+gs ds

Therefore, at critical points ‘t’

Ct=ft+g(t).

The critical numbers of C occur at the numbers t where Ct=ft+gt.

**Conclusion:**

The critical numbers of C occur at the numbers t where Ct=ft+gt.

#### To determine

**(b)**

The length of time T for Dt to equal the initial value of V.

#### Answer

T=30

#### Explanation

**1) Concept:**

Use The Fundamental Theorem of Calculus to find Dt.

**2) Theorem:**

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.

**3) Given:**

ft= V15-V450t if 0<t≤300 if t>30

gt=Vt212,900 t>0

**4) Calculations:**

Dt= ∫0tfs ds

By substituting value of fx,

Dt= ∫0tV15-V450sds

By integrating,

Dt=Vs15-s22·450t0

By applying limits,

Dt=Vt15-t2900

In order to determine the length of time T for Dt to equal the initial value of V

Let DT=V

D(T) Will be equal to V when t15-t2900=1

By solving the equation t15-t2900=1 for t, we have

900=60t-t2 Thus

t2-60t+900=0 Solving for t we have

t=30

The length of time T=30.

**Conclusion:**

T=30

#### To determine

**(c)**

The absolute minimum of C on (0, T].

#### Answer

The absolute minimum is C21.5=0.05472V

#### Explanation

**1) Given:**

Ct=1t∫0tfs+gs ds

From part (b) T=30.

**2) Calculations:**

Consider,

Ct=1t∫0tfs+gs ds

By substituting the values for ft and gt,

Ct=1t∫0tV15-V450s+V12900s2ds

On integrating,

Ct=1tV15s-V900S2+V38700S3t0

By applying limits,

Ct=1tV15t-V900t2+V38700t3

After simplifying,

Ct=V15-V900t+V38700t2

For critical values, set C't=0

-V900+V19350t=0

After solving for t,

t=19350900=21.5

Now,C21.5=V15-V90021.5+V3870021.52

After solving,

C21.5=0.05472V

Since t∈[0, 30]

C0=V15=0.066667V and C30=V15-V90030+V38700302=0.05659V

Observe that

C21.5<C30<C(0)

So the absolute minimum is C21.5=0.05472V

**Conclusion:**

The absolute minimum is C21.5=0.05472V

#### To determine

**(d)**

**To sketch:**

The graph for C and f+g

And verify the result in part(a)

#### Answer

Result is verified for part (a).

#### Explanation

**1) Given:**

Ct=V15-V900t+V38700t2

y=ft+g(t).

**2) Calculations:**

Notice that the critical point of Ct is at the intersection of ft+g(t) and Ct. This verifies part a)

**Final Solution:**

Thus, the result in part a) is verified and the graph is