The length of a box will be "1.8485 tt", width "1.8485 tt" and height "4.3899 tt".
According to the question,
Let,
→ [tex]V = x^2y[/tex]
[tex]x^2y = 15[/tex]
[tex]y = \frac{15}{x^2}[/tex]
Now,
→ [tex]Cost \ C(x) = Cost \ of \ top+Cost \ of \ bottom + Cost \ of \ side[/tex]
By substituting the values, we get
→ [tex]= 0.20 x^2+0.75 x^2+ 0.20(4xy)[/tex]
→ [tex]= 0.95x^2+0.20(4x \ \frac{15}{x^2} )[/tex]
→ [tex]= 0.95x^2+\frac{12}{x}[/tex]
→ [tex]C'(x) = 1.9 x-\frac{12}{x^2}[/tex]
[tex]0=1.9x-\frac{12}{x^2}[/tex]
→ [tex]C''(x) = 1.9+\frac{24}{x^3}[/tex]
[tex]1.9x = \frac{12}{x^2}[/tex]
hence,
→ [tex]x^3=\frac{12}{1.9}[/tex]
[tex]x = 3\sqrt{\frac{120}{19} }[/tex]
[tex]\approx 1.8485 \ tt[/tex]
and,
→ [tex]y = \frac{15}{(1.8485)^2}[/tex]
[tex]\approx 4.3899 \ tt[/tex]
Thus the above answer is appropriate.
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