Problem 1: The volume V and paper surface area of a conical paper cup are given by:
For V = 10 in 3 , compute the value of the radius, r that minimizes the area A. What is the corresponding value of the height, h? What is the minimum amount that r can vary from its optimal value before the area increases by 10%.
No products are associated with this question.
hi Suman Koirla, try this :
The Volume is given by : V=(1/3)*r²*h, and the surface A=pi*r*sqrt(r²+h²)
for V=10 m^3, we search for r that minimizes the Surface , :
Min(A) , SUBject to V=10
we have : h=3*V/pi*r² then : A=pi*r*sqrt(r²+90/pi²*r^4) .
Min(A) means the dA/dr=0=......=4*pi*r^3-180 /(2*sqrt(pi*r^4+90/r²))=0
Fast way to find R :
syms r A=(pi^2*r^2+90/r^2)^1/2 ezplot(A) S=subs(A,-6:0.1:6); % AXIS based on the first graph min(S)
1)so the minimum value for S=29.83 meter is R=1.89 ( FROM THE GRapH )
2) The corresponding value for h=3*10/(pi*1.89)=5.0525 meter .
Are you required to use a minimizer? The question can be solved analytically with a tiny amount of algebra together with some small calculus.
3)What is the minimum amount that r can vary from its optimal value before the area increases by 10% ( with fixed h ) :
Given S=29.83 m² and h=5.05 m, we have the new surface S2 :
__________ S2=S+0.1*S=32.81 m²=pi*r*\/ r²+h² .
S2²=pi².r^4 + pi²r²h² , make it as equation of 4th order :
r^4 + r² . h² -S2²/pi² = 0 ==> r^4 + 25.50 *r² - 109.7 = 0
We use the command "root" :
the Polynomial is a*r^4 + b*r^3 + c*r^2 + b*r + d = 0
a=1; b=0; c=25.50; d=-109.7
R_amount = roots([1 0 25.50 0 -109.7]) R_amount =
0.0000 + 5.4084i 0.0000 - 5.4084i 1.9366 -1.9366
The reasonable answer is the third one, R=1.9366 the amount change is
DELTA_R=1.9366-1.89=0.04 meter .