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DESCRIPTION
matlab
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1.- solución
Método punto
>> g=inline('(40-125/(x^(3)))^(1/2)');
>> ezplot('x',[0,2]),grid on, hold on
>> ezplot(g,[0,2])
>> x=g(x)
x = 6.278638092476449
>> x=g(x)
x = 6.284502692206101
>> x=g(x)
x = 6.284615073037374
>> x=g(x)
x = 6.284617222429021
>> x=g(x)
x = 6.284617263536701
>> x=g(x)
x = 6.284617264322895
>> x=g(x)
x = 6.284617264337932
>> x=g(x)
x = 6.284617264338219
>> x=g(x)
x = 6.284617264338225
>> x=g(x)
x = 6.284617264338225
el resultado es constante
Método newton
>> syms x
>> f=40-125/(x^(3))-x^(2);
>> ezplot(f,[0,2]),grid on
>> format long
>> x=0.5;
>> x=x-eval(f)/eval(diff(f))
x = 0.660068344724121
>> x=x-eval(f)/eval(diff(f))
x = 0.860197308107335
>> x=x-eval(f)/eval(diff(f))
x = 1.090186667003479
>> x=x-eval(f)/eval(diff(f))
x = 1.309185858834211
>> x=x-eval(f)/eval(diff(f))
x = 1.448512863258798
>> x=x-eval(f)/eval(diff(f))
x = 1.487727101258449
>> x=x-eval(f)/eval(diff(f))
x = 1.490099624826296
>> x=x-eval(f)/eval(diff(f))
x = 1.490107606835866
>> x=x-eval(f)/eval(diff(f))
x = 1.490107606925740
>> x=x-eval(f)/eval(diff(f))
x = 1.490107606925741
>> x=x-eval(f)/eval(diff(f))
x = 1.490107606925741
2.- solución
Método de punto
>> g=inline('(36594.38-(677.35/x)^(3))^(1/3)');
>> ezplot('x',[0,2]),grid on, hold on
>> ezplot(g,[0,2])
>> x=g(x)
x = 2.272527226990302e+02 + 3.936132618730813e+02i
>> x=g(x)
x = 33.201004576987586 + 0.000000000000000i
>> x=g(x)
x = 30.403091887427433 + 0.000000000000000i
>> x=g(x)
x = 29.447723749379470 + 0.000000000000000i
>> x=g(x)
x = 29.014086825233164 + 0.000000000000000i
>> x=g(x)
x = 28.793097913197517 + 0.000000000000000i
Luego de varias iteraciones mas:
>> x=g(x)
x = 28.520001374388325 + 0.000000000000000i
Método de newton
>> syms x
>> f=36594.38-(677.35/x)^(3)-x^(3);
>> ezplot(f,[0,2]),grid on
>> format long
>> x=23.7;
>> x=x-eval(f)/eval(diff(f))
x = 23.749986441637059
>> x=x-eval(f)/eval(diff(f))
x = 23.749998855440037
>> x=x-eval(f)/eval(diff(f))
x = 23.749998855479618
>> x=x-eval(f)/eval(diff(f))
x = 23.749998855479618
3.- solución:
Beneficio (y)
Venta=0.4x(30-x)
Costo =5+10lnx
Beneficio = venta –costo
Beneficio (y) = 0.4x(30-x)-( 5+10lnx)
Para hallar el beneficio max debemos obtener la derivada de la función benefio e igualarla a cero
0=12-0.8x-10/x
15-12.5/x=x
Método de punto:
>> g=inline('15-12.5/x');
>> ezplot('x',[0,2]),grid on, hold on
>> ezplot(g,[0,2])
>> x=14.11;
>> x=g(x)
x = 14.112903225806452
>> x=g(x)
x =14.114285714285714
>> x=g(x)
x = 14.114372469635628
>> x=g(x)
x = 14.114377913230548
luego de mas iteraciones:
>> x=g(x)
x = 14.114378277661476
>> x=g(x)
x = 14.114378277661476
Es el valor constante
Método de newton:
>> syms x
>> f= 12-0.8*x-10/x ;
>> ezplot(f,[0,2]),grid on
>> format long
>> x=x-eval(f)/eval(diff(f))
x = 0.885621606679308
>> x=x-eval(f)/eval(diff(f))
x = 0.885621722338507
>> x=x-eval(f)/eval(diff(f))
x = 0.885621722338523
>> x=x-eval(f)/eval(diff(f))
x = 0.885621722338524
4.- solución:
