USING MATLAB USING MATLAB The volume V and paper surface area A of a conical pap

Question

USING MATLAB

USING MATLAB

The volume V and paper surface area A of a conical paper cup are given by V = 1/3 pi r^2 h A = pi r squareroot r^2 + h^2 where r is the radius of the base of the cone and h is the height of the cone. a. By eliminating h, obtain the expression for A as a function of r and V. b. Create a user-defined function that accepts R as the only argument and computes A for a given value of V. Declare V to be global within the function. C. For V = 10 n.^3, use the function with the fminbnd function to compute the value of r that minimizes the area A. What is the corresponding value of the height h? Investigate the sensitivity of the solution by plotting V versus r. How much can R vary about its optimal value before the area increases 10 percent above its minimum value?

Explanation / Answer

solution 1 and 2

function Area=Cone(R)

V=10;

syms r h v A% all variables

eq1=v==1/3*pi*r^2*h;%eqution1

fprintf('h=1');

h_value=solve(eq1,h);%new h value for equation disp(h_value);

eq2=A==pi*r*sqrt(r^2+h_value^2);%equation 2 with new h value

fprintf('A=');

sol=solve(eq2,A);%solve equation 2 for A as a function of V and r

disp(sol);

solution=solve(eq2,v==V,r==R);

Area=double(solution.A);

end

Solution 3:

clc;

clear all;

v=10;

syms r h A % all variables

eq1=v==1/3*pi*r^2*h;%eqution 1

fprintf('h=');

h_value=solve(eq1,h);% new h value for eqution

disp(h_value);

eq2=A==pi*r*sqrt(r^2+h_value2);% eqution 2 with new h value

fprintf('A=');

sol=solve(eq2,A);% solve equation 2 for A as a function of V and r

func=matlabFuction(sol);

disp(func);

x1=0;

x2=10000000000;

x=fminbnd(func.x1,x2);

h_sol=solve(eq1,r==x);

fprintf('Min value of r=%f, New h value:%f ',x,double(h_sol.h));

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.