Answer-
For side length of 3.56 cm and height of 7.10 cm the cost will be minimum.
Solution-
Let us assume that,
x represents the length of the sides of the square base,
y represent the height.
Given the volume of the box is 90 cm³, so
[tex]\Rightarrow V=90\\\\\Rightarrow x^2\times y=90\\\\\Rightarrow y=\dfrac{90}{x^2}[/tex]
As the top and bottom cost $0.60 per cm² and the sides cost $0.30 per cm². Total cost C will be,
[tex]C=\text{cost for top and bottom}+\text{cost for rest 4 sides}\\\\=(2x^2\times 0.6)+(4xy\times 0.3)\\\\=(2x^2\times 0.6)+(4x\times \frac{90}{x^2}\times 0.3)\\\\=1.2x^2+ \dfrac{108}{x}[/tex]
Then,
[tex]C'=\dfrac{d}{dx}(1.2x^2+ \dfrac{108}{x})=2.4x-\dfrac{108}\\\\C''=\dfrac{d^2}{dx^2}(1.2x^2+ \dfrac{108}{x})=2.4+\dfrac{216}{x^3}[/tex]
As C'' has all positive terms so, for every positive value of x (as length can't be negative), C'' is positive.
So, for minima C' = 0
[tex]\Rightarrow 2.4x-\dfrac{108}{x^2}=0\\\\\Rightarrow 2.4x=\dfrac{108}{x^2}\\\\\Rightarrow x^3=\dfrac{108}{2.4}=45\\\\\Rightarrow x=3.56[/tex]
Then,
[tex]y=\dfrac{90}{x^2}[/tex]
[tex]y=\dfrac{90}{3.56^2}[/tex]
[tex]y=7.10[/tex]
Therefore, for side length of 3.56 cm and height of 7.10 cm the cost will be minimum.
