# Several Calculus Questions

*label*Mathematics

*timer*Asked: Jun 7th, 2018

*account_balance_wallet*$30

**Question description**

I have several calculus questions that I need help with. You MUST show your work and therefore must have a thorough understanding of calculus concepts and theories. You must NOT simply get answers from other online sites such as Coursehero etc. All work will be checked for accuracy so do not just copy answers you find online.

ANSWERS MUST BE TYPED!

## Tutor Answer

The solutions are ready. Please read them and ask if something is unclear, or if you want more explanations. etc.

1. β4π=1(β1)π (π + 11) = β12 + 13 β 14 + 15 = π.

2.

π₯2

π¦2

+ 11 = 1. 36 > 11 so it is a horizontal ellipse. The center is at the origin.

36

π = β36 = 6, π = β11, π = βπ2 β π 2 = 5, thus the foci are (βπ, π), (π, π).

2π¦ β π₯ = 5,

From the first equation π₯ = 2π¦ β 5, substitute it to the second:

π₯ + π¦ 2 β 25 = 0.

(2π¦ β 5)2 + π¦ 2 β 25 = 0, 5π¦ 2 β 20π¦ = 5π¦(π¦ β 4) = 0, π¦ = 0 ππ π¦ = 4.

So π₯ = β5 ππ π₯ = 3 and the answers are (βπ, π), (π, π).

3. {

2

4. lim(5π₯ β 3) = 5 lim π₯ β 3 = 5 β 5 β 3 = ππ.

π₯β5

π₯β5

4 β1 3 12

0 β17 β21 140

5. (1 4 6 β32) β |π
1 = π
1 β 4π
2 , π
3 = π
3 β 5π
2 | β (1

4

6

β32).

5 3 9 20

0 β17 β21 180

We see that the first and the second rows make a contradiction. So no solutions.

1 1

6. (1 β1

4 1

1 β5

1 1

1 β5

3 β1) β |π
2 = π
2 β π
1 , π
3 = π
3 β 4π
1 | β (0 β2 2

4 )β

0 β3 β3 18

1 β2

1 1 1 β5

1 1 1 β5

1

1

β |π
2 = β π
2 , π
3 = β π
3 | β (0 1 β1 β2) β |π
3 = π
3 β π
1 | β (0 1 β1 β2) β

2

3

0 1 1 β6

0 0 2 β4

1

1

1

β5

1

1

1

β5

1

β |π
3 = π
3 | β (0 1 β1 β2) β |π
2 = π
2 + π
3 | β (0 1 0 β4) β

2

0 0 1 β2

0 0 1 β2

1 0 0 1

β |π
1 = π
1 β π
2 β π
3 | β (0 1 0 β4), which means π = π, π = βπ, π = βπ.

0 0 1 β2

7. First, we know π₯ β₯ 0, π¦ β₯ 0. At least 336 calories means ππ + ππ β₯ πππ.

[these inequalities give the first quadrant except the triangle (0,42), (0,0), (48,0)]

8. π1 : 12 =

1β2

2

π2 : 12 + 42 =

, 1 = 1, which is true.

2β(24β6β1)

2

π3 : 12 + 42 + 72 =

, 17 = 17, which is true.

3β(6β9β9β1)

2

, 66 =

3β44

2

= 66, which is true.

9. ππ : βππ=1 π(π + 1) =

π(π+1)(π+2)

(π+1)(π+2)(π+3)

3

3

, so ππ+1 : βπ+1

π=1 π(π + 1) =

.

[I am not sure what is "simplify ππ+1 "]

10. Denote π₯ the number of pounds of breakfast tea and π¦ of afternoon tea.

Then the constraints are:

1

1

2

1

π₯ β₯ 0, π¦ β₯ 0,

π₯ + π¦ β€ 45,

π₯ + π¦ β€ 70.

3

2

3

2

The profit functions ...

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