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# Thread: Math geniuses I need your help again with this optimization problem gtfih (1.7k reps)

1. ## Math geniuses I need your help again with this optimization problem gtfih (1.7k reps)

A rectangle is constructed with its base on the diameter of a semicircle with radius 5 cm and with two vertices on the semicircle. What are the dimensions of the rectangle with maximum area?

I'm guessing the picture is just a regular rectangle with one side being attached to a semicircle.

I'm having problems with figuring out what/how is my constraint going to look like? First I wrote it in terms of the perimeter on the semicircle, then the rectangle, and now I just have no idea. I wrote the area function as A(x) = xy.

Has anyone done this one before? The answer in the book says 10/root2 cm by 5/root2 cm.

2. Originally Posted by E-PRO
A rectangle is constructed with its base on the diameter of a semicircle with radius 5 cm and with two vertices on the semicircle. What are the dimensions of the rectangle with maximum area?

I'm guessing the picture is just a regular rectangle with one side being attached to a semicircle.

I'm having problems with figuring out what/how is my constraint going to look like? First I wrote it in terms of the perimeter on the semicircle, then the rectangle, and now I just have no idea. I wrote the area function as A(x) = xy.

Has anyone done this one before? The answer in the book says 10/root2 cm by 5/root2 cm.
probably not much help but wouldnt the area be something along the lines of A(x)=[((pieR^2)/2)+xy? to take into account the half circle. [(pieR^2)/2] being the area of a semicircle
apologies if it isn't much help, been awhile since I did this

3. ~<=3

4. Originally Posted by Willie1993
probably not much help but wouldnt the area be something along the lines of A(x)=[((pieR^2)/2)+xy? to take into account the half circle. [(pieR^2)/2] being the area of a semicircle
apologies if it isn't much help, been awhile since I did this
I didn't think of that but I'll try it. I suck at optimization problems that don't clearly give you the constraint. Like this other one says that the perimeter of 3 fences is 200 meters, that's easy, but this one just gives you the radius of a semicircle attached to a rectangle, wtf cuz...

5. currently doing this problem in calculus

saving spot and typing my answer

actually idk but the hint i can give u is the area of the semi circle is [5 pie] idk about the rectangle tho x should be 10 but the question doesn't seem clear enough to discover the Y. All of my problems specifically say the x and y boundaries are bound by blah blah blah

6. Originally Posted by E-PRO
I didn't think of that but I'll try it. I suck at optimization problems that don't clearly give you the constraint. Like this other one says that the perimeter of 3 fences is 200 meters, that's easy, but this one just gives you the radius of a semicircle attached to a rectangle, wtf cuz...
atleast when youre given the radius of the circle you now know that the top and bottom of the rectangle is 10cm, but does it mention anything about a height limit? elsewise I dont see why you couldn't say that the height is 10, or 100, or 1000, or inifite

7. Originally Posted by Willie1993
atleast when youre given the radius of the circle you now know that the top and bottom of the rectangle is 10cm, but does it mention anything about a height limit? elsewise I dont see why you couldn't say that the height is 10, or 100, or 1000, or inifite
That's what I was thinking but apparently not lol.

8. Originally Posted by Willie1993
atleast when youre given the radius of the circle you now know that the top and bottom of the rectangle is 10cm, but does it mention anything about a height limit? elsewise I dont see why you couldn't say that the height is 10, or 100, or 1000, or inifite
also brah, a quick google search gave me the following link

the question 19 seems to be identical to your problem, take a look at it, might help

hope I could help

9. 3.50 foots

10. let x and y be the sides of the rectangle
its easy to see that (r)^2 = x^2 + y^2 eq(1) where r is the radius of the semicircle
the area of the retacngle is A=xy eq(2)

we know that y=sqrt(r^2)-x^2) from(1)

(1)->(2)
A=x*sqrt(r^2)-x^2)
A=sqrt(x^2(r^2)-x^4)
sqrt(x^2(r^2)-x^4)
A is now a A(x) , A is max when sqrt(x^2(r)^2)-x^4) is max and sqrt(x^2(r^2)-x^4) is max when x^2(r^2)-x^4 is max

f(x)=x^2(r^2)-x^4
f'(x)=2(r^2)x -4x^3 =0 (we want max value)
-4x^2=-2(r^2)
x^2=2(r^2)/3=2(r^2)/3
x=sqrt(2/4)*r=sqrt(2)*r/2
wich is the max value of side x...now u get max area.
if you are a calculus student justification is a bit more complex , you'd have to analize the graph of the function A(x) and justify that the value x=sqrt(2)*r/2 is not just a local maximum (it has to be a global maximum , but remember the problems restrictions.)

edit erros...

11. Originally Posted by Willie1993
also brah, a quick google search gave me the following link

the question 19 seems to be identical to your problem, take a look at it, might help

hope I could help
Fuark how is it that I couldn't find that when I searched... Well thanks man, remind me to rep you again.

12. Originally Posted by E-PRO
A rectangle is constructed with its base on the diameter of a semicircle with radius 5 cm and with two vertices on the semicircle. What are the dimensions of the rectangle with maximum area?

I'm guessing the picture is just a regular rectangle with one side being attached to a semicircle.

I'm having problems with figuring out what/how is my constraint going to look like? First I wrote it in terms of the perimeter on the semicircle, then the rectangle, and now I just have no idea. I wrote the area function as A(x) = xy.

Has anyone done this one before? The answer in the book says 10/root2 cm by 5/root2 cm.
Whenever they are asking for the max/min of something, it's all about setting the derivative to 0.

What you need to do is find the derivative of A(x) = 2*x*y, set it equal to 0, and solve for x. Perhaps, you're confused because there are two variables?

No worries. Write y in terms of x. Equation of a circle is x^2+y^2 = 25 --> y = sqrt(25-x^2). Great! Now, we can take the derivative of A(x) = 2x(25-x^2)^(1/2).

When you set d[A(x)]/dx = 0, and solve for x, you get x = 5/sqrt(2). Substitute that back into y = sqrt(25-x^2), and you also get y = 5/sqrt(2). To bet the base, you multiply x by 2 since the rectangle will span two quadrants. Now you get 10/sqrt(2) by 5/sqrt(2) with an area of 25.

13. petition to add a math keys to normal key boards. It hurts my brain to read anything in this damn thread

14. Originally Posted by ppham27
Whenever they are asking for the max/min of something, it's all about setting the derivative to 0.

What you need to do is find the derivative of A(x) = 2*x*y, set it equal to 0, and solve for x. Perhaps, you're confused because there are two variables?

No worries. Write y in terms of x. Equation of a circle is x^2+y^2 = 25 --> y = sqrt(25-x^2). Great! Now, we can take the derivative of A(x) = 2x(25-x^2)^(1/2).

When you set d[A(x)]/dx = 0, and solve for x, you get x = 5/sqrt(2). Substitute that back into y = sqrt(25-x^2), and you also get y = 5/sqrt(2). To bet the base, you multiply x by 2 since the rectangle will span two quadrants. Now you get 10/sqrt(2) by 5/sqrt(2) with an area of 25.
I know all about derivatives, critical values, concavity, second derivatives, etc and I know how to set up and solve optimization problems and related rates but I have a problem with coming up with the equations for most optimization problems.

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