Mixing
Norman window problem with area provided
Clash Royale CLAN TAG#URR8PPP
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I've seen plenty of norman window threads online and on this platform as well, though I'm stumped on a slightly different version of the usual question. Generally a perimeter is provided and we are asked to optimise the area. I understand the solutions presented to this type of question but I've been given the following problem and have been so far unable to solve it.
Determine the smallest possible perimeter if the area of the window is fixed at 6.8 square metres.
I've developed a formula for both area and perimeter and have isolated the variable 'h': h=(13.6-Àr2)/2r.
After substituting that into 2h+r+Àh, which was my formula for perimeter, I am left with f(r)=(68+5r2)/5r. I took the derivative and got f'(r)=(5r2-68)/5r2. But this still doesn't help me out, as the function never reaches zero. I'm totally bewildered. I'm not sure if I've gone about this the right way so I'll leave a drawing of the problem and the variables and my working here.
Please excuse the terrible drawing. Thanks
Update: I accidentally treated r as the diameter in some of my calculations but the end result remains the same, the formula for perimeter cannot be proven correctly with the results obtained and the derivative's domain cannot equal 0.
calculus geometry derivatives
edited Jul 22 at 10:48
Aretino
21.7k21342
asked Jul 22 at 9:05
afisc123
12
add a comment |Â
up vote
0
down vote
favorite
I've seen plenty of norman window threads online and on this platform as well, though I'm stumped on a slightly different version of the usual question. Generally a perimeter is provided and we are asked to optimise the area. I understand the solutions presented to this type of question but I've been given the following problem and have been so far unable to solve it.
Determine the smallest possible perimeter if the area of the window is fixed at 6.8 square metres.
I've developed a formula for both area and perimeter and have isolated the variable 'h': h=(13.6-Àr2)/2r.
After substituting that into 2h+r+Àh, which was my formula for perimeter, I am left with f(r)=(68+5r2)/5r. I took the derivative and got f'(r)=(5r2-68)/5r2. But this still doesn't help me out, as the function never reaches zero. I'm totally bewildered. I'm not sure if I've gone about this the right way so I'll leave a drawing of the problem and the variables and my working here.
Please excuse the terrible drawing. Thanks
Update: I accidentally treated r as the diameter in some of my calculations but the end result remains the same, the formula for perimeter cannot be proven correctly with the results obtained and the derivative's domain cannot equal 0.
calculus geometry derivatives
edited Jul 22 at 10:48
Aretino
21.7k21342
asked Jul 22 at 9:05
afisc123
12
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 9:11
2
Who is Norman Window?
– Christian Blatter
Jul 22 at 9:46
1
$5r^2-68=0implies r=sqrt68/5$. What's wrong with that?
– Aretino
Jul 22 at 10:52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've seen plenty of norman window threads online and on this platform as well, though I'm stumped on a slightly different version of the usual question. Generally a perimeter is provided and we are asked to optimise the area. I understand the solutions presented to this type of question but I've been given the following problem and have been so far unable to solve it.
Determine the smallest possible perimeter if the area of the window is fixed at 6.8 square metres.
I've developed a formula for both area and perimeter and have isolated the variable 'h': h=(13.6-Àr2)/2r.
After substituting that into 2h+r+Àh, which was my formula for perimeter, I am left with f(r)=(68+5r2)/5r. I took the derivative and got f'(r)=(5r2-68)/5r2. But this still doesn't help me out, as the function never reaches zero. I'm totally bewildered. I'm not sure if I've gone about this the right way so I'll leave a drawing of the problem and the variables and my working here.
Please excuse the terrible drawing. Thanks
Update: I accidentally treated r as the diameter in some of my calculations but the end result remains the same, the formula for perimeter cannot be proven correctly with the results obtained and the derivative's domain cannot equal 0.
calculus geometry derivatives
edited Jul 22 at 10:48
Aretino
21.7k21342
asked Jul 22 at 9:05
afisc123
12
I've seen plenty of norman window threads online and on this platform as well, though I'm stumped on a slightly different version of the usual question. Generally a perimeter is provided and we are asked to optimise the area. I understand the solutions presented to this type of question but I've been given the following problem and have been so far unable to solve it.
Determine the smallest possible perimeter if the area of the window is fixed at 6.8 square metres.
I've developed a formula for both area and perimeter and have isolated the variable 'h': h=(13.6-Àr2)/2r.
After substituting that into 2h+r+Àh, which was my formula for perimeter, I am left with f(r)=(68+5r2)/5r. I took the derivative and got f'(r)=(5r2-68)/5r2. But this still doesn't help me out, as the function never reaches zero. I'm totally bewildered. I'm not sure if I've gone about this the right way so I'll leave a drawing of the problem and the variables and my working here.
Please excuse the terrible drawing. Thanks
Update: I accidentally treated r as the diameter in some of my calculations but the end result remains the same, the formula for perimeter cannot be proven correctly with the results obtained and the derivative's domain cannot equal 0.
calculus geometry derivatives
edited Jul 22 at 10:48
Aretino
21.7k21342
asked Jul 22 at 9:05
afisc123
12
edited Jul 22 at 10:48
Aretino
21.7k21342
edited Jul 22 at 10:48
Aretino
21.7k21342
edited Jul 22 at 10:48
Aretino
21.7k21342
21.7k21342
asked Jul 22 at 9:05
afisc123
12
asked Jul 22 at 9:05
afisc123
12
asked Jul 22 at 9:05
afisc123
12
12
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 9:11
2
Who is Norman Window?
– Christian Blatter
Jul 22 at 9:46
1
$5r^2-68=0implies r=sqrt68/5$. What's wrong with that?
– Aretino
Jul 22 at 10:52
add a comment |Â
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 9:11
2
Who is Norman Window?
– Christian Blatter
Jul 22 at 9:46
1
$5r^2-68=0implies r=sqrt68/5$. What's wrong with that?
– Aretino
Jul 22 at 10:52
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 9:11
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 9:11
2
2
Who is Norman Window?
– Christian Blatter
Jul 22 at 9:46
Who is Norman Window?
– Christian Blatter
Jul 22 at 9:46
1
1
$5r^2-68=0implies r=sqrt68/5$. What's wrong with that?
– Aretino
Jul 22 at 10:52
$5r^2-68=0implies r=sqrt68/5$. What's wrong with that?
– Aretino
Jul 22 at 10:52
add a comment |Â
1 Answer
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oldest
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0
down vote
After googling "Norman window", it appears to be a rectangle surmounted by a semicircle.
Hint
With $r$ being the radius of the semicircle and $h$ being the height of the rectangle, we can write the perimeter $P$ and area $A$ as
$$P = pi r + 2h + 2r$$
$$A= fracpi r^22 + 2rh = 6.8$$
From the equation for area we find that $$h = frac6.82r - fracpi r4$$
Inserting this into the perimeter equation we find
$$P=r(fracpi2+2)+frac6.8r$$
Now differentiate this equation, set it to zero, find $r$ and you are done.
answered Jul 22 at 10:57
Jens
3,0652826
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
After googling "Norman window", it appears to be a rectangle surmounted by a semicircle.
Hint
With $r$ being the radius of the semicircle and $h$ being the height of the rectangle, we can write the perimeter $P$ and area $A$ as
$$P = pi r + 2h + 2r$$
$$A= fracpi r^22 + 2rh = 6.8$$
From the equation for area we find that $$h = frac6.82r - fracpi r4$$
Inserting this into the perimeter equation we find
$$P=r(fracpi2+2)+frac6.8r$$
Now differentiate this equation, set it to zero, find $r$ and you are done.
answered Jul 22 at 10:57
Jens
3,0652826
add a comment |Â
up vote
0
down vote
After googling "Norman window", it appears to be a rectangle surmounted by a semicircle.
Hint
With $r$ being the radius of the semicircle and $h$ being the height of the rectangle, we can write the perimeter $P$ and area $A$ as
$$P = pi r + 2h + 2r$$
$$A= fracpi r^22 + 2rh = 6.8$$
From the equation for area we find that $$h = frac6.82r - fracpi r4$$
Inserting this into the perimeter equation we find
$$P=r(fracpi2+2)+frac6.8r$$
Now differentiate this equation, set it to zero, find $r$ and you are done.
answered Jul 22 at 10:57
Jens
3,0652826
add a comment |Â
up vote
0
down vote
up vote
0
down vote
After googling "Norman window", it appears to be a rectangle surmounted by a semicircle.
Hint
With $r$ being the radius of the semicircle and $h$ being the height of the rectangle, we can write the perimeter $P$ and area $A$ as
$$P = pi r + 2h + 2r$$
$$A= fracpi r^22 + 2rh = 6.8$$
From the equation for area we find that $$h = frac6.82r - fracpi r4$$
Inserting this into the perimeter equation we find
$$P=r(fracpi2+2)+frac6.8r$$
Now differentiate this equation, set it to zero, find $r$ and you are done.
answered Jul 22 at 10:57
Jens
3,0652826
After googling "Norman window", it appears to be a rectangle surmounted by a semicircle.
Hint
With $r$ being the radius of the semicircle and $h$ being the height of the rectangle, we can write the perimeter $P$ and area $A$ as
$$P = pi r + 2h + 2r$$
$$A= fracpi r^22 + 2rh = 6.8$$
From the equation for area we find that $$h = frac6.82r - fracpi r4$$
Inserting this into the perimeter equation we find
$$P=r(fracpi2+2)+frac6.8r$$
Now differentiate this equation, set it to zero, find $r$ and you are done.
answered Jul 22 at 10:57
Jens
3,0652826
answered Jul 22 at 10:57
Jens
3,0652826
answered Jul 22 at 10:57
Jens
3,0652826
answered Jul 22 at 10:57
Jens
3,0652826
3,0652826
add a comment |Â
add a comment |Â
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1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 9:11
2
Who is Norman Window?
– Christian Blatter
Jul 22 at 9:46
1
$5r^2-68=0implies r=sqrt68/5$. What's wrong with that?
– Aretino
Jul 22 at 10:52