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Find the edges size of a regular triangular pyramid with a given volume, which has the smallest sum of edges.

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0

$begingroup$

$a\$ - edge length at the base of the pyramid.

$b\$ - length of the side edges of the pyramid.

$V\$ = $const\$.

I minimize: $3a+3b->min\$

$V=frac13Sh\$

$h=sqrtb^2-frac13a^2\$

$S=fracsqrt34a^2\$

$V=frac13*fracsqrt34a^2*sqrtb^2-frac13a^2\$ - from here I expressed the parameter $b\$ and substituted in $3a+3b->min\$

$b=sqrtfracV^2+frac3a^6432frac3a^4144\$

After substitution, I found the derivative and equated it to zero, but the derivative was too complex, it seems to me that I am doing something wrong.

Derivative:
$3+fraca^6-288V^2a^5*sqrtfrac48V^2a^4+fraca^23=0\$

How to express $a\$ from here?

optimization triangle

share|cite|improve this question

edited Mar 14 at 15:45

bvl

asked Mar 14 at 15:11

bvlbvl

165

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get the formula for $h$?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 15:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the Pythagorean theorem: $h^2=b^2-(frac23*fracsqrt3a2)^2\$
  $endgroup$
  – bvl
  Mar 14 at 15:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, i get $$b=sqrtfrac144V^23a^4+frac13a^2$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 15:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get this too. $b=sqrtfracV^2+frac3a^6432frac3a^4144 = sqrtfrac144V^23a^4+fraca^23\$
  $endgroup$
  – bvl
  Mar 14 at 15:36
  

  
  
  
  

add a comment | 

0

$begingroup$

$a\$ - edge length at the base of the pyramid.

$b\$ - length of the side edges of the pyramid.

$V\$ = $const\$.

I minimize: $3a+3b->min\$

$V=frac13Sh\$

$h=sqrtb^2-frac13a^2\$

$S=fracsqrt34a^2\$

$V=frac13*fracsqrt34a^2*sqrtb^2-frac13a^2\$ - from here I expressed the parameter $b\$ and substituted in $3a+3b->min\$

$b=sqrtfracV^2+frac3a^6432frac3a^4144\$

After substitution, I found the derivative and equated it to zero, but the derivative was too complex, it seems to me that I am doing something wrong.

Derivative:
$3+fraca^6-288V^2a^5*sqrtfrac48V^2a^4+fraca^23=0\$

How to express $a\$ from here?

optimization triangle

share|cite|improve this question

edited Mar 14 at 15:45

bvl

asked Mar 14 at 15:11

bvlbvl

165

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get the formula for $h$?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 15:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the Pythagorean theorem: $h^2=b^2-(frac23*fracsqrt3a2)^2\$
  $endgroup$
  – bvl
  Mar 14 at 15:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, i get $$b=sqrtfrac144V^23a^4+frac13a^2$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 15:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get this too. $b=sqrtfracV^2+frac3a^6432frac3a^4144 = sqrtfrac144V^23a^4+fraca^23\$
  $endgroup$
  – bvl
  Mar 14 at 15:36
  

  
  
  
  

add a comment | 

$begingroup$

$a\$ - edge length at the base of the pyramid.

$b\$ - length of the side edges of the pyramid.

$V\$ = $const\$.

I minimize: $3a+3b->min\$

$V=frac13Sh\$

$h=sqrtb^2-frac13a^2\$

$S=fracsqrt34a^2\$

$V=frac13*fracsqrt34a^2*sqrtb^2-frac13a^2\$ - from here I expressed the parameter $b\$ and substituted in $3a+3b->min\$

$b=sqrtfracV^2+frac3a^6432frac3a^4144\$

After substitution, I found the derivative and equated it to zero, but the derivative was too complex, it seems to me that I am doing something wrong.

Derivative:
$3+fraca^6-288V^2a^5*sqrtfrac48V^2a^4+fraca^23=0\$

How to express $a\$ from here?

optimization triangle

share|cite|improve this question

edited Mar 14 at 15:45

bvl

asked Mar 14 at 15:11

bvlbvl

165

$endgroup$

share|cite|improve this question

edited Mar 14 at 15:45

bvl

asked Mar 14 at 15:11

bvlbvl

165

share|cite|improve this question

edited Mar 14 at 15:45

bvl

asked Mar 14 at 15:11

bvlbvl

165

asked Mar 14 at 15:11

bvlbvl

165

asked Mar 14 at 15:11

bvlbvl

165

1 Answer
1

active

oldest

votes

0

$begingroup$

As brought up in the comments, we have that

$$
b=sqrtfrac144V^23a^4+frac13a^2.
$$

You wish to minimize the function

$$
f(a)=a+sqrtfrac144V^23a^4+frac13a^2
$$

by equating the derivative to zero (note that I dropped the scaling multiplication by $3$, since it does not affect the minimizer). Introducing the notation $k=144V^2/3$ and $m=1/3$, the derivative of $f$ is

$$
f'(a)=1-fracbig(frac2ka^5-mabig)sqrtfracka^4+ma^2
$$

Equate this to zero, and get

$$
sqrtfracka^4+ma^2=frac2ka^5-ma
$$

Multiply both sides by $a^5$:

$$
sqrtka^6+ma^12=2k-ma^6.
$$

Introduce the notation $xequiva^6$, and this becomes

$$
sqrtmx^2+kx=2k-mx.
$$

Square both sides (at this point we need to be careful about choosing the right signs):

$$
m(1-m)x^2+k(1+4m)x-4k^2=0.
$$

Solve with the quadratic formula (choose the proper sign), and substitute $x=a^6$ to get the final answer.

share|cite|improve this answer

answered Mar 14 at 16:40

David M.David M.

2,058418

$endgroup$

add a comment | 

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0

$begingroup$

As brought up in the comments, we have that

$$
b=sqrtfrac144V^23a^4+frac13a^2.
$$

You wish to minimize the function

$$
f(a)=a+sqrtfrac144V^23a^4+frac13a^2
$$

by equating the derivative to zero (note that I dropped the scaling multiplication by $3$, since it does not affect the minimizer). Introducing the notation $k=144V^2/3$ and $m=1/3$, the derivative of $f$ is

$$
f'(a)=1-fracbig(frac2ka^5-mabig)sqrtfracka^4+ma^2
$$

Equate this to zero, and get

$$
sqrtfracka^4+ma^2=frac2ka^5-ma
$$

Multiply both sides by $a^5$:

$$
sqrtka^6+ma^12=2k-ma^6.
$$

Introduce the notation $xequiva^6$, and this becomes

$$
sqrtmx^2+kx=2k-mx.
$$

Square both sides (at this point we need to be careful about choosing the right signs):

$$
m(1-m)x^2+k(1+4m)x-4k^2=0.
$$

Solve with the quadratic formula (choose the proper sign), and substitute $x=a^6$ to get the final answer.

share|cite|improve this answer

answered Mar 14 at 16:40

David M.David M.

2,058418

$endgroup$

add a comment | 

0

$begingroup$

As brought up in the comments, we have that

$$
b=sqrtfrac144V^23a^4+frac13a^2.
$$

You wish to minimize the function

$$
f(a)=a+sqrtfrac144V^23a^4+frac13a^2
$$

by equating the derivative to zero (note that I dropped the scaling multiplication by $3$, since it does not affect the minimizer). Introducing the notation $k=144V^2/3$ and $m=1/3$, the derivative of $f$ is

$$
f'(a)=1-fracbig(frac2ka^5-mabig)sqrtfracka^4+ma^2
$$

Equate this to zero, and get

$$
sqrtfracka^4+ma^2=frac2ka^5-ma
$$

Multiply both sides by $a^5$:

$$
sqrtka^6+ma^12=2k-ma^6.
$$

Introduce the notation $xequiva^6$, and this becomes

$$
sqrtmx^2+kx=2k-mx.
$$

Square both sides (at this point we need to be careful about choosing the right signs):

$$
m(1-m)x^2+k(1+4m)x-4k^2=0.
$$

Solve with the quadratic formula (choose the proper sign), and substitute $x=a^6$ to get the final answer.

share|cite|improve this answer

answered Mar 14 at 16:40

David M.David M.

2,058418

$endgroup$

add a comment | 

$begingroup$

As brought up in the comments, we have that

$$
b=sqrtfrac144V^23a^4+frac13a^2.
$$

You wish to minimize the function

$$
f(a)=a+sqrtfrac144V^23a^4+frac13a^2
$$

by equating the derivative to zero (note that I dropped the scaling multiplication by $3$, since it does not affect the minimizer). Introducing the notation $k=144V^2/3$ and $m=1/3$, the derivative of $f$ is

$$
f'(a)=1-fracbig(frac2ka^5-mabig)sqrtfracka^4+ma^2
$$

Equate this to zero, and get

$$
sqrtfracka^4+ma^2=frac2ka^5-ma
$$

Multiply both sides by $a^5$:

$$
sqrtka^6+ma^12=2k-ma^6.
$$

Introduce the notation $xequiva^6$, and this becomes

$$
sqrtmx^2+kx=2k-mx.
$$

Square both sides (at this point we need to be careful about choosing the right signs):

$$
m(1-m)x^2+k(1+4m)x-4k^2=0.
$$

Solve with the quadratic formula (choose the proper sign), and substitute $x=a^6$ to get the final answer.

share|cite|improve this answer

answered Mar 14 at 16:40

David M.David M.

2,058418

$endgroup$

share|cite|improve this answer

answered Mar 14 at 16:40

David M.David M.

2,058418

answered Mar 14 at 16:40

David M.David M.

2,058418

answered Mar 14 at 16:40

David M.David M.

2,058418