a closed rectangular box whose base is twice as long as it is wide has a volume of 36000 cm^3.the material for the top cost 10 centavo per sq cm,that for the sides and bottom cost 5 centavos per sq cm.find the dimensions that will make the cost of making the box a minimum

Answer 1

x(2x)(z) = 36000
so, the height z is 18000/x^2

the area is 4x^2 + 2xz + 4xz
the cost in centavos is
c = 10*4x^2 + 5*2xz + 5*4xz
= 40x^2 + 30xz
= 40x^2 +540000/x
= 40(x^2 + 13500/x)

minimum cost is when dc/dx=0
dc/dx = 40(2x-13500/x^2)
= 80(x^3-6750)/x^2
dc/dx=0 when x^3 = 6750, or x = 15∛2

so the box is 15∛2 by 30∛2 by 40∛2

Answer 2

The question said that the bottom's price is the same with the sides price. so basically you need to divide the top and bottom which is 20(2x)(x)+10(2x)(x)