Optimization problem (rectangle inscribed below parabola) - possible textbook mistake












0












$begingroup$


I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.



The problem:




A rectangle is located below a parabola, which is given by $$y = 3x- frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.



c) What is that area?




Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.



enter image description here



If that's helpful, the textbook's answer for b) and c) are $3-sqrt{3} times 3$ and $9 + 9sqrt{3}$ respectively.



Thank you.






calculus optimization







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  • 1




    $begingroup$
    $base=2sqrt{3}$
    $endgroup$
    – Aleksas Domarkas
    Jan 6 at 8:19
















0












$begingroup$


I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.



The problem:




A rectangle is located below a parabola, which is given by $$y = 3x- frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.



c) What is that area?




Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.



enter image description here



If that's helpful, the textbook's answer for b) and c) are $3-sqrt{3} times 3$ and $9 + 9sqrt{3}$ respectively.



Thank you.






calculus optimization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $base=2sqrt{3}$
    $endgroup$
    – Aleksas Domarkas
    Jan 6 at 8:19














0












0








0





$begingroup$


I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.



The problem:




A rectangle is located below a parabola, which is given by $$y = 3x- frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.



c) What is that area?




Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.



enter image description here



If that's helpful, the textbook's answer for b) and c) are $3-sqrt{3} times 3$ and $9 + 9sqrt{3}$ respectively.



Thank you.






calculus optimization







share|cite|improve this question











$endgroup$




I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.



The problem:




A rectangle is located below a parabola, which is given by $$y = 3x- frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.



c) What is that area?




Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.



enter image description here



If that's helpful, the textbook's answer for b) and c) are $3-sqrt{3} times 3$ and $9 + 9sqrt{3}$ respectively.



Thank you.






calculus optimization




calculus optimization






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 8:11







bru1987

















asked Jan 6 at 8:00









bru1987bru1987

1,0691123




1,0691123








  • 1




    $begingroup$
    $base=2sqrt{3}$
    $endgroup$
    – Aleksas Domarkas
    Jan 6 at 8:19














  • 1




    $begingroup$
    $base=2sqrt{3}$
    $endgroup$
    – Aleksas Domarkas
    Jan 6 at 8:19








1




1




$begingroup$
$base=2sqrt{3}$
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:19




$begingroup$
$base=2sqrt{3}$
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:19










1 Answer
1






active

oldest

votes


















1












$begingroup$

You made a slight error when trying to find the base:



$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$



You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.



Addition: From here, using $h = 3$ as you found, you get



$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$



Try plugging in $c = 3-sqrt{3}$ in $f(c)$:



$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$



Through confirmation, you can see this point coincides with the local maximum.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
    $endgroup$
    – bru1987
    Jan 6 at 9:16










  • $begingroup$
    Yes, that’s the correct answer!
    $endgroup$
    – KM101
    Jan 6 at 9:24










  • $begingroup$
    thank you and have a great day =)
    $endgroup$
    – bru1987
    Jan 6 at 9:26






  • 1




    $begingroup$
    Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
    $endgroup$
    – KM101
    Jan 6 at 9:31













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

You made a slight error when trying to find the base:



$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$



You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.



Addition: From here, using $h = 3$ as you found, you get



$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$



Try plugging in $c = 3-sqrt{3}$ in $f(c)$:



$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$



Through confirmation, you can see this point coincides with the local maximum.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
    $endgroup$
    – bru1987
    Jan 6 at 9:16










  • $begingroup$
    Yes, that’s the correct answer!
    $endgroup$
    – KM101
    Jan 6 at 9:24










  • $begingroup$
    thank you and have a great day =)
    $endgroup$
    – bru1987
    Jan 6 at 9:26






  • 1




    $begingroup$
    Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
    $endgroup$
    – KM101
    Jan 6 at 9:31


















1












$begingroup$

You made a slight error when trying to find the base:



$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$



You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.



Addition: From here, using $h = 3$ as you found, you get



$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$



Try plugging in $c = 3-sqrt{3}$ in $f(c)$:



$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$



Through confirmation, you can see this point coincides with the local maximum.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
    $endgroup$
    – bru1987
    Jan 6 at 9:16










  • $begingroup$
    Yes, that’s the correct answer!
    $endgroup$
    – KM101
    Jan 6 at 9:24










  • $begingroup$
    thank you and have a great day =)
    $endgroup$
    – bru1987
    Jan 6 at 9:26






  • 1




    $begingroup$
    Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
    $endgroup$
    – KM101
    Jan 6 at 9:31
















1












1








1





$begingroup$

You made a slight error when trying to find the base:



$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$



You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.



Addition: From here, using $h = 3$ as you found, you get



$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$



Try plugging in $c = 3-sqrt{3}$ in $f(c)$:



$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$



Through confirmation, you can see this point coincides with the local maximum.






share|cite|improve this answer











$endgroup$



You made a slight error when trying to find the base:



$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$



You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.



Addition: From here, using $h = 3$ as you found, you get



$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$



Try plugging in $c = 3-sqrt{3}$ in $f(c)$:



$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$



Through confirmation, you can see this point coincides with the local maximum.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 6 at 9:36

























answered Jan 6 at 8:38









KM101KM101

6,0701525




6,0701525












  • $begingroup$
    thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
    $endgroup$
    – bru1987
    Jan 6 at 9:16










  • $begingroup$
    Yes, that’s the correct answer!
    $endgroup$
    – KM101
    Jan 6 at 9:24










  • $begingroup$
    thank you and have a great day =)
    $endgroup$
    – bru1987
    Jan 6 at 9:26






  • 1




    $begingroup$
    Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
    $endgroup$
    – KM101
    Jan 6 at 9:31




















  • $begingroup$
    thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
    $endgroup$
    – bru1987
    Jan 6 at 9:16










  • $begingroup$
    Yes, that’s the correct answer!
    $endgroup$
    – KM101
    Jan 6 at 9:24










  • $begingroup$
    thank you and have a great day =)
    $endgroup$
    – bru1987
    Jan 6 at 9:26






  • 1




    $begingroup$
    Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
    $endgroup$
    – KM101
    Jan 6 at 9:31


















$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16




$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16












$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24




$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24












$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26




$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26




1




1




$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31






$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31




















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