Need help with math problems please him?

1. A flower bed is to be shape of a circular sector of radius r and central angle θ. Find r and θ if the area is fixed and the perimeter is a minimum.



2. Suppose that the position of a particle on a straight line at time t is s(t) = at - (1 + a^4)t^2. Show that the particle moves forward initially when a is positive but ultimately retreats. Show also that for different values of a the maximum possible distance that particle can move forward is 1/8.



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Need help with math problems please him?
#1



The area of the sector is:

A = r2 Θ/2



The perimeter of the sector is:

P(r,Θ) = 2r+rΘ



Solving the first for Θ gives:

Θ = 2A / r2



Substituting gives:

P(r) = 2r + 2A/r



Differentiating gives:

P'(r) = 2 - 2A/r2



Solving for critical points gives:

0 = 2 - 2A/r2

r2 = A

r = √A



This gives:

Θ = 2A / √(A)2 = 2

P = 2√A + 2A/√A = 4√A



#2



Take the derivative:

s'(t) = a - 2(1+a^4)t



This is a line with negative slope, but a positive intercept. That means the particle's speed is positive initially, but then goes negative. Since s(0)=0, we know it starts at the point 0, moves forward for a while, and then goes backwards.



The distance the particle moves forward depends on when it turns around. First, find that critical point:



s'(t) = 0

a - 2(1+a^4)t = 0

a = 2(1+a^4)t

t = a / 2(1+a^4)



Plug that into the function s(t), to see how far it goes:



s(a /(2+2a^4))



= a(a / (2+2a^4)) - (1 + a^4)(a / (2+2a^4))2



= a2 / (2+2a^4) - a2 / 2(2+2a^4)



= a2 / 2(2+2a^4)



This is a function of a that we want to maximize:

M(a) = a2 / 2(2+2a^4)



Differentiate using the quotient rule

M'(a) = a(a-a^4) / 2(1+a^4)2



Solve M'(a) = 0



a(a-a^4) / 2(1+a^4)2 = 0

(a-a^4) = 0

a = a^4

a = ±1

(or a = 0, since we divided by a)



a = 1 is the appropriate solution (since a%26gt;0)



M(1) = 12 / 2(2+2×1^4) = 1/8



That is the maximum distance forward
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