Calculus Homework

Given the function g(x)=4×3+6×2−72xg(x)=4×3+6×2-72x, find the first derivative, g'(x)g′(x).
g'(x)=g′(x)=

Notice that g'(x)=0g′(x)=0 when x=−3x=-3, that is, g'(−3)=0g′(-3)=0.

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Now, we want to know whether there is a local minimum or local maximum at x=−3x=-3, so we will use the second derivative test.
Find the second derivative, g”(x)g′′(x).
g”(x)=g′′(x)=

Evaluate g”(−3)g′′(-3).
g”(−3)=g′′(-3)=

Based on the sign of this number, does this mean the graph of g(x)g(x) is concave up or concave down at x=−3x=-3?
At x=−3x=-3 the graph of g(x)g(x) is

Based on the concavity of g(x)g(x) at x=−3x=-3, does this mean that there is a local minimum or local maximum at x=−3x=-3?
At x=−3x=-3 there is a local

Question 1. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

The function f(x)=−2×3+30×2−126x+10f(x)=-2×3+30×2-126x+10 has one local minimum and one local maximum.
This function has a local minimum at xx =
with value

and a local maximum at xx =
with value

Question 2. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

The function f(x)=2×3−36×2+210x+5f(x)=2×3-36×2+210x+5 has one local minimum and one local maximum.
This function has a local minimum at xx =
with function value

and a local maximum at xx =
with function value

Question 3. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

A box with a square base and open top must have a volume of 131072 cm3cm3. We wish to find the dimensions of the box that minimize the amount of material used.

First, find a formula for the surface area of the box in terms of only xx, the length of one side of the square base.
[Hint: use the volume formula to express the height of the box in terms of xx.]
Simplify your formula as much as possible.
A(x)=A(x)=

Next, find the derivative, A'(x)A′(x).
A'(x)=A′(x)=

Now, calculate when the derivative equals zero, that is, when A'(x)=0A′(x)=0. [Hint: multiply both sides by x2x2.]
A'(x)=0A′(x)=0 when x=x=

We next have to make sure that this value of xx gives a minimum value for the surface area. Let’s use the second derivative test. Find A”(x)(x).
A”(x)=(x)=

Evaluate A”(x)(x) at the xx-value you gave above.

NOTE: Since your last answer is positive, this means that the graph of A(x)A(x) is concave up around that value, so the zero of A'(x)A′(x) must indicate a local minimum for A(x)A(x). (Your boss is happy now.)

Question 4. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=11−x2y=11-x2. What are the dimensions of such a rectangle with the greatest possible area?

Width =
Height =

Question 5. Last Attempt:  0 out of 1
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A piece of cardboard measuring 8 inches by 12 inches is formed into an open-top box by cutting squares with side length xx from each corner and folding up the sides.

Find a formula for the volume of the box in terms of xx

V(x)V(x) =

Find the value for xx that will maximize the volume of the box

xx =

Question 6. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

A cylinder shaped can needs to be constructed to hold 300 cubic centimeters of soup. The material for the sides of the can costs 0.03 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.05 cents per square centimeter. Find the dimensions for the can that will minimize production cost.

Helpful information:
h : height of can, r : radius of can

Volume of a cylinder: V=πr2hV=πr2h

Area of the sides: A=2πrhA=2πrh

Area of the top/bottom: A=πr2A=πr2

To minimize the cost of the can:
Radius of the can:
Height of the can:
Minimum cost:     cents

Question 7. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 450 cubic centimeters of soup. The sides and bottom of the container will be made of styrofoam costing 0.03 cents per square centimeter. The top will be made of glued paper, costing 0.05 cents per square centimeter. Find the dimensions for the package that will minimize production cost.

Helpful information:
h : height of cylinder, r : radius of cylinder

Volume of a cylinder: V=πr2hV=πr2h

Area of the sides: A=2πrhA=2πrh

Area of the top/bottom: A=πr2A=πr2

To minimize the cost of the package:
Radius:     cm
Height:     cm
Minimum cost:     cents

Question 8. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

The function f(x)=2×3−45×2+300x+2f(x)=2×3-45×2+300x+2 has one local minimum and one local maximum.
This function has a local minimum at xx equals

with value

and a local maximum at xx equals

with value

Question 9. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

Given the function g(x)=6×3−36×2+54xg(x)=6×3-36×2+54x, find the first derivative, g'(x)g′(x).
g'(x)=g′(x)=

Notice that g'(x)=0g′(x)=0 when x=3x=3, that is, g'(3)=0g′(3)=0.

Now, we want to know whether there is a local minimum or local maximum at x=3x=3, so we will use the second derivative test.
Find the second derivative, g”(x)(x).
g”(x)=(x)=

Evaluate g”(3)(3).
g”(3)=(3)=

Based on the sign of this number, does this mean the graph of g(x)g(x) is concave up or concave down at x=3x=3?
[Answer either up or down — watch your spelling!!]
At x=3x=3 the graph of g(x)g(x) is concave

Based on the concavity of g(x)g(x) at x=3x=3, does this mean that there is a local minimum or local maximum at x=3x=3?
[Answer either minimum or maximum — watch your spelling!!]
At x=3x=3 there is a local

Question 10. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

Open-box Problem. An open-box (top open) is made from a rectangular material of dimensions a=12a=12 inches by b=11b=11 inches by cutting a square of side xx at each corner and turning up the sides (see the figure). Determine the value of xx that results in a box the maximum volume.

12345678910111213-1123456789101112-1xab[Graphs generated by this script: setBorder(5); initPicture(-1,12+1,-1,11+1);axes(1,1,1,1,1);fill=”yellow”;path([[0,0],[0,11],[12,11],[12,0],[0,0]]);fill=”cyan”;path([[0,0],[0,11/6],[11/6,11/6],[11/6,0],[0,0]]);fill=”cyan”;path([[12-11/6,0],[12-11/6,11/6],[12,11/6],[12,0],[12-11/6,0]]);fill=”cyan”;path([[12-11/6,11-11/6],[12-11/6,11],[12,11],[12,11-11/6],[12-11/6,11-11/6]]);fill=”cyan”;path([[0,11-11/6],[0,11],[11/6,11],[11/6,11-11/6],[0,11-11/6]]);text([12+.5,11-11/12],”x”);text([12/2,11+11/12],”a”);text([12+.5,11-11/2],”b”);]

Following the steps to solve the problem. Check Show Answer only after you have tried hard.

(1) Express the volume VV as a function of xx: V=V=

(2) Determine the domain of the function VV of xx (in interval form):

(3) Expand the function VV for easier differentiation: V=V=

(4) Find the derivative of the function VV: V’=V′=

(5) Find the critical point(s) in the domain of VV:

(6) The value of VV at the left endpoint is

(7) The value of VV at the right endpoint is

(8) The maximum volume is V=V=

(9) Answer the original question. The value of xx that maximizes the volume is:

Question 11. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

The base and sides of a container is made of wood panels. The container does not have a lid. The base and sides are rectangular. The width of the container is xx cmcm . The length is fourth times the width. The volume of the container is 392cm3392cm3 .

Determine the minimum surface area that this container will have.

 

Question 12. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

The function f(x)=2×3−33×2+180x+3f(x)=2×3-33×2+180x+3 has one local minimum and one local maximum.
This function has a local minimum at xx equals

with value

and a local maximum at xx equals

with value

Question 13. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

Given the function g(x)=6×3+9×2−36xg(x)=6×3+9×2-36x, find the first derivative, g'(x)g′(x).
g'(x)=g′(x)=

Notice that g'(x)=0g′(x)=0 when x=1x=1, that is, g'(1)=0g′(1)=0.

Now, we want to know whether there is a local minimum or local maximum at x=1x=1, so we will use the second derivative test.
Find the second derivative, g”(x)(x).
g”(x)=(x)=

Evaluate g”(1)(1).
g”(1)=(1)=

Based on the sign of this number, does this mean the graph of g(x)g(x) is concave up or concave down at x=1x=1?
[Answer either up or down — watch your spelling!!]
At x=1x=1 the graph of g(x)g(x) is concave

Based on the concavity of g(x)g(x) at x=1x=1, does this mean that there is a local minimum or local maximum at x=1x=1?
[Answer either minimum or maximum — watch your spelling!!]
At x=1x=1 there is a local

Question 14. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

Open-box Problem. An open-box (top open) is made from a rectangular material of dimensions a=15a=15 inches by b=15b=15 inches by cutting a square of side xx at each corner and turning up the sides (see the figure). Determine the value of xx that results in a box the maximum volume.

12345678910111213141516-112345678910111213141516-1xab[Graphs generated by this script: setBorder(5); initPicture(-1,15+1,-1,15+1);axes(1,1,1,1,1);fill=”yellow”;path([[0,0],[0,15],[15,15],[15,0],[0,0]]);fill=”cyan”;path([[0,0],[0,15/6],[15/6,15/6],[15/6,0],[0,0]]);fill=”cyan”;path([[15-15/6,0],[15-15/6,15/6],[15,15/6],[15,0],[15-15/6,0]]);fill=”cyan”;path([[15-15/6,15-15/6],[15-15/6,15],[15,15],[15,15-15/6],[15-15/6,15-15/6]]);fill=”cyan”;path([[0,15-15/6],[0,15],[15/6,15],[15/6,15-15/6],[0,15-15/6]]);text([15+.5,15-15/12],”x”);text([15/2,15+15/12],”a”);text([15+.5,15-15/2],”b”);]

Following the steps to solve the problem. Check Show Answer only after you have tried hard.

(1) Express the volume VV as a function of xx: V=V=

(2) Determine the domain of the function VV of xx (in interval form):

(3) Expand the function VV for easier differentiation: V=V=

(4) Find the derivative of the function VV: V’=V′=

(5) Find the critical point(s) in the domain of VV:

(6) The value of VV at the left endpoint is

(7) The value of VV at the right endpoint is

(8) The maximum volume is V=V=

(9) Answer the original question. The value of xx that maximizes the volume is:

Question 15. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

The base and sides of a container is made of wood panels. The container does not have a lid. The base and sides are rectangular. The width of the container is xx cmcm . The length is three times the width. The volume of the container is 150cm3150cm3 .

Determine the minimum surface area that this container will have.

 

Question 16. Last Attempt:  0 out of 1
Score in Gradebook:   0 out of 1

 

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