Calculus Parcel Optimisation Problem

Clash Royale CLAN TAG#URR8PPP

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The question is worded as follows:

"A courier firm has restrictions on the size of parcels that it will carry at the normal rate. The length of a parcel must be no less than twice its width. The sum of the length, width and height of any parcel also must be no more than 136 millimetres."

I thought I had solved this problem, I wrote an equation for the Volume, as well as a function graphing the width. Turns out I was way off and totally messed it up. I just can't understand how to solve this as the only way I can express $l$ seems to be with an inequality and I can't solve that. When I solved it first I tried: $l = 2w - x$

But I don't think that's correct. Any help would be appreciated. I'm also asked to provide an assumption which is made about the shape of the package.
Thanks.

algebra-precalculus geometry derivatives optimization

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edited Jul 28 at 11:01

caverac

11k2927

asked Jul 28 at 10:49

afisc123

12

add a comment | 

up vote
0
down vote

favorite

The question is worded as follows:

"A courier firm has restrictions on the size of parcels that it will carry at the normal rate. The length of a parcel must be no less than twice its width. The sum of the length, width and height of any parcel also must be no more than 136 millimetres."

I thought I had solved this problem, I wrote an equation for the Volume, as well as a function graphing the width. Turns out I was way off and totally messed it up. I just can't understand how to solve this as the only way I can express $l$ seems to be with an inequality and I can't solve that. When I solved it first I tried: $l = 2w - x$

But I don't think that's correct. Any help would be appreciated. I'm also asked to provide an assumption which is made about the shape of the package.
Thanks.

algebra-precalculus geometry derivatives optimization

share|cite|improve this question

edited Jul 28 at 11:01

caverac

11k2927

asked Jul 28 at 10:49

afisc123

12

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

The question is worded as follows:

"A courier firm has restrictions on the size of parcels that it will carry at the normal rate. The length of a parcel must be no less than twice its width. The sum of the length, width and height of any parcel also must be no more than 136 millimetres."

I thought I had solved this problem, I wrote an equation for the Volume, as well as a function graphing the width. Turns out I was way off and totally messed it up. I just can't understand how to solve this as the only way I can express $l$ seems to be with an inequality and I can't solve that. When I solved it first I tried: $l = 2w - x$

But I don't think that's correct. Any help would be appreciated. I'm also asked to provide an assumption which is made about the shape of the package.
Thanks.

algebra-precalculus geometry derivatives optimization

share|cite|improve this question

edited Jul 28 at 11:01

caverac

11k2927

asked Jul 28 at 10:49

afisc123

12

The question is worded as follows:

"A courier firm has restrictions on the size of parcels that it will carry at the normal rate. The length of a parcel must be no less than twice its width. The sum of the length, width and height of any parcel also must be no more than 136 millimetres."

I thought I had solved this problem, I wrote an equation for the Volume, as well as a function graphing the width. Turns out I was way off and totally messed it up. I just can't understand how to solve this as the only way I can express $l$ seems to be with an inequality and I can't solve that. When I solved it first I tried: $l = 2w - x$

But I don't think that's correct. Any help would be appreciated. I'm also asked to provide an assumption which is made about the shape of the package.
Thanks.

algebra-precalculus geometry derivatives optimization

share|cite|improve this question

edited Jul 28 at 11:01

caverac

11k2927

asked Jul 28 at 10:49

afisc123

12

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 11:01

caverac

11k2927

edited Jul 28 at 11:01

caverac

11k2927

edited Jul 28 at 11:01

caverac

11k2927

11k2927

asked Jul 28 at 10:49

afisc123

12

asked Jul 28 at 10:49

afisc123

12

asked Jul 28 at 10:49

afisc123

12

12

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote

Calling

$$
x = mboxlength\
y = mboxwidth\
z = mboxheight
$$
the restrictions can be established as

$$
x ge 0\
y ge 0\
z ge 0\
xge 2 y\
x+y + z le 136
$$

resulting in the restriction volume attached

so the maximal feasible dimensions volume in one piece is

$$
x = frac5449\
y = frac2729\
z = frac1363
$$

Assuming that the maximum volume fits tight we can proceed as follows

$$
V = x y z\
x = 2yRightarrow V = 2y^2z\
x+y+z = LRightarrow 3y + z = LRightarrow z = L - 3y\
V = 2y^2(L-3y)
$$

so the extremum points obey

$$
fracddyV(y) = 4yL-18y^2=0Rightarrow y = frac2729
$$

etc.

share|cite|improve this answer

edited Jul 28 at 12:17

answered Jul 28 at 11:12

Cesareo

5,6412412

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks but unfortunately it has to be done at a year 11 standard, using only a graphing calculator and algebraic working. The question is part of a calculus/differentiation unit so the answer will be attainable through finding an equation and solving its derivative equal to zero. The problem I'm having is how to deal with the inequalities as they don't allow for plotting as a function on the graphical calculator.
  – afisc123
  Jul 28 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @afisc123 Attached a note I hope it helps.
  – Cesareo
  Jul 28 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much, it did.
  – afisc123
  Jul 29 at 3:49
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Calling

$$
x = mboxlength\
y = mboxwidth\
z = mboxheight
$$
the restrictions can be established as

$$
x ge 0\
y ge 0\
z ge 0\
xge 2 y\
x+y + z le 136
$$

resulting in the restriction volume attached

so the maximal feasible dimensions volume in one piece is

$$
x = frac5449\
y = frac2729\
z = frac1363
$$

Assuming that the maximum volume fits tight we can proceed as follows

$$
V = x y z\
x = 2yRightarrow V = 2y^2z\
x+y+z = LRightarrow 3y + z = LRightarrow z = L - 3y\
V = 2y^2(L-3y)
$$

so the extremum points obey

$$
fracddyV(y) = 4yL-18y^2=0Rightarrow y = frac2729
$$

etc.

share|cite|improve this answer

edited Jul 28 at 12:17

answered Jul 28 at 11:12

Cesareo

5,6412412

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks but unfortunately it has to be done at a year 11 standard, using only a graphing calculator and algebraic working. The question is part of a calculus/differentiation unit so the answer will be attainable through finding an equation and solving its derivative equal to zero. The problem I'm having is how to deal with the inequalities as they don't allow for plotting as a function on the graphical calculator.
  – afisc123
  Jul 28 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @afisc123 Attached a note I hope it helps.
  – Cesareo
  Jul 28 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much, it did.
  – afisc123
  Jul 29 at 3:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

Calling

$$
x = mboxlength\
y = mboxwidth\
z = mboxheight
$$
the restrictions can be established as

$$
x ge 0\
y ge 0\
z ge 0\
xge 2 y\
x+y + z le 136
$$

resulting in the restriction volume attached

so the maximal feasible dimensions volume in one piece is

$$
x = frac5449\
y = frac2729\
z = frac1363
$$

Assuming that the maximum volume fits tight we can proceed as follows

$$
V = x y z\
x = 2yRightarrow V = 2y^2z\
x+y+z = LRightarrow 3y + z = LRightarrow z = L - 3y\
V = 2y^2(L-3y)
$$

so the extremum points obey

$$
fracddyV(y) = 4yL-18y^2=0Rightarrow y = frac2729
$$

etc.

share|cite|improve this answer

edited Jul 28 at 12:17

answered Jul 28 at 11:12

Cesareo

5,6412412

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks but unfortunately it has to be done at a year 11 standard, using only a graphing calculator and algebraic working. The question is part of a calculus/differentiation unit so the answer will be attainable through finding an equation and solving its derivative equal to zero. The problem I'm having is how to deal with the inequalities as they don't allow for plotting as a function on the graphical calculator.
  – afisc123
  Jul 28 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @afisc123 Attached a note I hope it helps.
  – Cesareo
  Jul 28 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much, it did.
  – afisc123
  Jul 29 at 3:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Calling

$$
x = mboxlength\
y = mboxwidth\
z = mboxheight
$$
the restrictions can be established as

$$
x ge 0\
y ge 0\
z ge 0\
xge 2 y\
x+y + z le 136
$$

resulting in the restriction volume attached

so the maximal feasible dimensions volume in one piece is

$$
x = frac5449\
y = frac2729\
z = frac1363
$$

Assuming that the maximum volume fits tight we can proceed as follows

$$
V = x y z\
x = 2yRightarrow V = 2y^2z\
x+y+z = LRightarrow 3y + z = LRightarrow z = L - 3y\
V = 2y^2(L-3y)
$$

so the extremum points obey

$$
fracddyV(y) = 4yL-18y^2=0Rightarrow y = frac2729
$$

etc.

share|cite|improve this answer

edited Jul 28 at 12:17

answered Jul 28 at 11:12

Cesareo

5,6412412

Calling

$$
x = mboxlength\
y = mboxwidth\
z = mboxheight
$$
the restrictions can be established as

$$
x ge 0\
y ge 0\
z ge 0\
xge 2 y\
x+y + z le 136
$$

resulting in the restriction volume attached

so the maximal feasible dimensions volume in one piece is

$$
x = frac5449\
y = frac2729\
z = frac1363
$$

Assuming that the maximum volume fits tight we can proceed as follows

$$
V = x y z\
x = 2yRightarrow V = 2y^2z\
x+y+z = LRightarrow 3y + z = LRightarrow z = L - 3y\
V = 2y^2(L-3y)
$$

so the extremum points obey

$$
fracddyV(y) = 4yL-18y^2=0Rightarrow y = frac2729
$$

etc.

share|cite|improve this answer

edited Jul 28 at 12:17

answered Jul 28 at 11:12

Cesareo

5,6412412

share|cite|improve this answer

share|cite|improve this answer

edited Jul 28 at 12:17

edited Jul 28 at 12:17

edited Jul 28 at 12:17

answered Jul 28 at 11:12

Cesareo

5,6412412

answered Jul 28 at 11:12

Cesareo

5,6412412

answered Jul 28 at 11:12

Cesareo

5,6412412

5,6412412

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks but unfortunately it has to be done at a year 11 standard, using only a graphing calculator and algebraic working. The question is part of a calculus/differentiation unit so the answer will be attainable through finding an equation and solving its derivative equal to zero. The problem I'm having is how to deal with the inequalities as they don't allow for plotting as a function on the graphical calculator.
  – afisc123
  Jul 28 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @afisc123 Attached a note I hope it helps.
  – Cesareo
  Jul 28 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much, it did.
  – afisc123
  Jul 29 at 3:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks but unfortunately it has to be done at a year 11 standard, using only a graphing calculator and algebraic working. The question is part of a calculus/differentiation unit so the answer will be attainable through finding an equation and solving its derivative equal to zero. The problem I'm having is how to deal with the inequalities as they don't allow for plotting as a function on the graphical calculator.
  – afisc123
  Jul 28 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @afisc123 Attached a note I hope it helps.
  – Cesareo
  Jul 28 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much, it did.
  – afisc123
  Jul 29 at 3:49
  

  
  
  
  

Thanks but unfortunately it has to be done at a year 11 standard, using only a graphing calculator and algebraic working. The question is part of a calculus/differentiation unit so the answer will be attainable through finding an equation and solving its derivative equal to zero. The problem I'm having is how to deal with the inequalities as they don't allow for plotting as a function on the graphical calculator.
– afisc123
Jul 28 at 11:49

Thanks but unfortunately it has to be done at a year 11 standard, using only a graphing calculator and algebraic working. The question is part of a calculus/differentiation unit so the answer will be attainable through finding an equation and solving its derivative equal to zero. The problem I'm having is how to deal with the inequalities as they don't allow for plotting as a function on the graphical calculator.
– afisc123
Jul 28 at 11:49

@afisc123 Attached a note I hope it helps.
– Cesareo
Jul 28 at 12:18

@afisc123 Attached a note I hope it helps.
– Cesareo
Jul 28 at 12:18

Thanks so much, it did.
– afisc123
Jul 29 at 3:49

Thanks so much, it did.
– afisc123
Jul 29 at 3:49

add a comment | 

 

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