Solving a cubic equation, it is found that the dimensions of the container are of 1 inch, 9 inches and 5 inches.
The volume of the cubic container is given by the following equation:
[tex]V(h) = 2h^3 - 9h^2 + 4h[/tex]
Since it holds 45 cubic inches of merchandise, we have that [tex]V(h) = 45[/tex], then:
[tex]V(h) = 2h^3 - 9h^2 + 4h[/tex]
[tex]45 = 2h^3 - 9h^2 + 4h[/tex]
[tex]2h^3 - 9h^2 + 4h - 45 = 0[/tex]
Using a cubic equation calculator, the real solution is [tex]h = 5[/tex], thus, the dimensions are:
[tex]h = 5[/tex]
[tex]h - 4 = 5 - 4 = 1[/tex]
[tex]2h - 1 = 2(5) - 1 = 10 - 1 = 9[/tex]
The dimensions of the container are of 1 inch, 9 inches and 5 inches.
A similar problem is given at https://brainly.com/question/16606259
