Respuesta :
Answer:
(a)Length =2 feet
(b)Width =2 feet
(c)Height=3 feet
Step-by-step explanation:
Let the dimensions of the box be x, y and z
The rectangular box has a square base.
Therefore, Volume of the box[tex]V=x^2z[/tex]
Volume of the box[tex]=12 ft^3\\[/tex]
[tex]Therefore, x^2z=12\\z=\frac{12}{x^2}[/tex]
The material for the base costs [tex]\$0.17/ft^2[/tex], the material for the sides costs [tex]\$0.10/ft^2[/tex], and the material for the top costs [tex]\$0.13/ft^2[/tex].
Area of the base [tex]=x^2[/tex]
Cost of the Base [tex]=\$0.17x^2[/tex]
Area of the sides [tex]=4xz[/tex]
Cost of the sides=[tex]=\$0.10(4xz)[/tex]
Area of the Top [tex]=x^2[/tex]
Cost of the Base [tex]=\$0.13x^2[/tex]
Total Cost, [tex]C(x,z) =0.17x^2+0.13x^2+0.10(4xz)[/tex]
Substituting [tex]z=\frac{12}{x^2}[/tex]
[tex]C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}[/tex]
To minimize C(x), we solve for the derivative and obtain its critical point
[tex]C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2[/tex]
Recall: [tex]z=\frac{12}{x^2}=\frac{12}{2^2}=3\\[/tex]
Therefore, the dimensions that minimizes the cost of the box are:
(a)Length =2 feet
(b)Width =2 feet
(c)Height=3 feet
