Respuesta :
Answer:
[tex]C(x)=\dfrac{20x^3+1715000}{x}\\$Minimum cost, C(35)=\$29,400[/tex]
The dimensions that will lead to minimum cost of the box are a base length of 35 cm and a height of 140 cm.
Step-by-step explanation:
Volume of the Square-Based box=171,500 cubic cm
Let the length of a side of the base=x cm
Volume [tex]=x^2h[/tex]
[tex]x^2h=171,500\\h=\dfrac{171500}{x^2}[/tex]
The material for the top and bottom of the box costs $10.00 per square centimeter.
Surface Area of the Top and Bottom [tex]=2x^2[/tex]
Therefore, Cost of the Top and Bottom [tex]=\$10X2x^2=20x^2[/tex]
The material for the sides costs $2.50 per square centimeter.
Surface Area of the Sides=4xh
Cost of the sides=$2.50 X 4xh =10xh
[tex]\text{Substitute h}$=\dfrac{171500}{x^2} $into 10xh\\Cost of the sides=10x(\dfrac{171500}{x^2})=\dfrac{1715000}{x}[/tex]
Therefore, total Cost of the box
[tex]= 20x^2+\dfrac{1715000}{x}\\C(x)=\dfrac{20x^3+1715000}{x}[/tex]
To find the minimum total cost, we solve for the critical points of C(x). This is obtained by equating its derivative to zero and solving for x.
[tex]C'(x)=\dfrac{40x^3-1715000}{x^2}\\\dfrac{40x^3-1715000}{x^2}=0\\40x^3-1715000=0\\40x^3=1715000\\x^3=1715000\div 40\\x^3=42875\\x=\sqrt[3]{42875}=35[/tex]
Recall that:
[tex]h=\dfrac{171500}{x^2}\\Therefore:\\h=\dfrac{171500}{35^2}=140cm[/tex]
The dimensions that will lead to minimum costs are base length of 35cm and height of 140cm.
Therefore, the minimum total cost, at x=35cm
[tex]C(35)=\dfrac{20(35)^3+1715000}{35}=\$29,400[/tex]
