Answer:
The height of the container that minimize cost is 3.08 cm
Step-by-step explanation:
Let
h ---> the height of the container
we know that
The volume of the box is equal to
[tex]V=(3x)(4x)h[/tex]
[tex]V=12x^2h[/tex]
[tex]V=48\ cm^3[/tex]
substitute
[tex]48=12x^2h[/tex]
[tex]h=\frac{4}{x^2}[/tex]
The function cost is equal to
[tex]C=3.50(2)(12x^2)+4.40(14x)(h)[/tex]
[tex]C=84x^2+61.6x\frac{4}{x^2}[/tex]
[tex]C=84x^2+\frac{246.4}{x}[/tex]
To find out the minimum cost determine the first derivative
[tex]\frac{dC}{dx}=168x-\frac{246.4}{x^2}[/tex]
equate to zero
[tex]168x=\frac{246.4}{x^2}\\x^3=1.4667\\x=1.14\ cm[/tex]
Find the height of the container
[tex]h=\frac{4}{x^2}[/tex]
substitute the value of x
[tex]h=\frac{4}{1.14^2}=3.08\ cm[/tex]
