Respuesta :
Answer:
Base Side Length =1.59m
Height = 3.16 m
Step-by-step explanation:
Volume of the box = [tex]8$ m^3[/tex]
Let the base dimensions = x and y
Let the height of the box =h
However, for any optimal configuration, Width = Length as varying the length and width to be other than equal reduces the volume for the same total(w+l)
Volume, V=xyh=8
Since x=y
[tex]h=\dfrac{8}{x^2}[/tex]
Surface Area of the box
[tex]= 2(x^2+xh+xh)\\=2x^2+4xh[/tex]
The material for the top and bottom costs twice as much as the material for the sides.
Let the cost of the sides =$1 per square meter
Cost of the material for the sides = 4xh
Cost of the material for the top and bottom = [tex]=2*2x^2=4x^2[/tex]
Therefore:
Total Cost, [tex]C= 4xh+4x^2[/tex]
Substitution of [tex]h=\frac{8}{x^2}[/tex] into C
[tex]C= 4x(\frac{8}{x^2})+4x^2\\=\frac{32}{x}+4x^2\\C(x)=\dfrac{4x^3+32}{x}[/tex]
To minimize C(x), we find its derivative and solve for the critical points.
[tex]C'(x)=\dfrac{8x^3-32}{x^2}\\$Setting C'(x) to zero\\8x^3-32=0\\8x^3=32\\x^3=4\\x=\sqrt[3]{4}\\ x=1.59$ m[/tex]
To verify if it is a minimum, we use the second derivative test
[tex]C''(x)=8+\frac{64}{x^3}\\C''(1.59)=23.92[/tex]
Since C''(x) is greater than zero, it is a minimum point.
Recall:
[tex]h=\frac{8}{x^2}\\h=\frac{8}{1.59^2}=3.16$ m[/tex]
Therefore, the dimensions that minimizes the cost are:
Base Side Lengths of 1.59m; and
Height of 3.16 m
