Respuesta :
Answer:
[tex]Total cost=10x^{2} dollars+\frac{168dollars}{x}[/tex]
Explanation:
First at all, let's see the figure in the attachment where we define:
Width is "x", Lenght must be "2x" and Height is "h".
The volume of a rectangular storage container is defined as:
V=lengh*width*height
So, replacing values, we have:
V=[tex]2x^{2} h[/tex]
Considering that V=14m3 we can clear "h" in function of "x" (the width):
[tex]V=2x^{2} h=14; h=\frac{7}{x^{2}}[/tex]
Now, we calculate the areas of the container:
[tex]Ax_{1} =x*h\\Ax_{2}=2x*h\\Ax_{3}=2x*x[/tex]
Where: Ax1 is the side 1 area; Ax2 is the side 2 area and Ax3 is the base area
Replacing "h" on the previous equations, we have:
[tex]Ax_{1}=x*h=x*\frac{7}{x^{2} }=\frac{7}{x}\\Ax_{2}=2x*h=2x*\frac{7}{x^{2} }=\frac{14}{x}\\Ax_{3}=x*2x=2x^{2}[/tex]
Remember that the container is open at the top, so we have to calculate just one area in the base. The sides 1 and 2 are 2 of each one.
So, we have: Total area = 2*Ax1 + 2*Ax2 + Ax3
Now for the total cost of materials, we have: Total cost=Cost (2*Ax1) + Cost (2*Ax2) + Cost (Ax3)
For the sides 1 and 2, we have a cost of:
[tex]Cost Ax_{1}=\frac{7}{x}m^{2} *\frac{4 dollars}{m^{2} }=\frac{28dollars}{x}\\Cost Ax_{2}=\frac{14}{x}m^{2} *\frac{4 dollars}{m^{2} }=\frac{56dollars}{x}\\Cost Ax_{3}=(2x^{2})m^{2}*\frac{5 dollars}{m^{2}}=10x^{2}dollars\\[/tex]
Finally, total cost is:
[tex]Total cost=2*\frac{28 dollars}{x} + 2*\frac{56 dollars}{x} + 10x^{2} dollars\\Total cost=10x^{2} dollars+\frac{168dollars}{x}[/tex]
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