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I have several questions all pertaining to the same problem,different sections. This is the problem and following that I will have my work.
Twenty feet of wire is to be used to form two figures. In each of the following cases, how much wire should be used for each figure so that the total enclosed area is maximum?
a.) Equilateral Triangle and Square
b.) Square and Regular Pentagon
c.) Regular Pentagon and Regular Hexagon
d.) Regular Hexagon and Circle
a.) A=x^2+3^(1/2)s^2/4 20=4x+3s
Then I solved for s s=(20-4x)/3
that left me with the following equation
A=x^2+(3^(1/2)/4)((20-4x)/3)^2
Took the derivative and got A'=(2x(9+4*3^(1/2)))/9-(40*3^(1/2))/9)
solved for x x=2.17482. Then solved for s s=3.76691
The square uses 8.69928 ft of wire, and the triangle uses 11.3007 ft. of wire.
Is this correct?
b,c, and d I've had some trouble with so here is some of the work I did, can anyone help further.
b.) A=s^2+(5/4)(cot (pie/5))(x^2) 20=4s+5x
i solved for s s=(20-5x)/4
Now is where I start having difficulty
I substitute for s to get A=((20-5x)/4)^2+(5/4)(cot (pie/5))x^2
Now I need to get the derivative. Did I get the area equation correct? and if so, my derivative is coming out as A'=(50x-200)+(10x/4)(cot(pie/5))
That just doesn't seem right to me.
c and d are pretty much the same set up. It is the cot(pie/n) that is messing me up I think. I know the only difference on d is the area of a circle rather then a regular polygon. Thank you for all your help in advance!![/list]
Twenty feet of wire is to be used to form two figures. In each of the following cases, how much wire should be used for each figure so that the total enclosed area is maximum?
a.) Equilateral Triangle and Square
b.) Square and Regular Pentagon
c.) Regular Pentagon and Regular Hexagon
d.) Regular Hexagon and Circle
a.) A=x^2+3^(1/2)s^2/4 20=4x+3s
Then I solved for s s=(20-4x)/3
that left me with the following equation
A=x^2+(3^(1/2)/4)((20-4x)/3)^2
Took the derivative and got A'=(2x(9+4*3^(1/2)))/9-(40*3^(1/2))/9)
solved for x x=2.17482. Then solved for s s=3.76691
The square uses 8.69928 ft of wire, and the triangle uses 11.3007 ft. of wire.
Is this correct?
b,c, and d I've had some trouble with so here is some of the work I did, can anyone help further.
b.) A=s^2+(5/4)(cot (pie/5))(x^2) 20=4s+5x
i solved for s s=(20-5x)/4
Now is where I start having difficulty
I substitute for s to get A=((20-5x)/4)^2+(5/4)(cot (pie/5))x^2
Now I need to get the derivative. Did I get the area equation correct? and if so, my derivative is coming out as A'=(50x-200)+(10x/4)(cot(pie/5))
That just doesn't seem right to me.
c and d are pretty much the same set up. It is the cot(pie/n) that is messing me up I think. I know the only difference on d is the area of a circle rather then a regular polygon. Thank you for all your help in advance!![/list]
