[ASK] Minimum Surface Area

In summary, the volume of a cuboid box with a square base is 2 litres. The production cost per unit of its top and bottom is twice the production cost per unit of its lateral sides. The minimum production cost is reached when the surface area is 8 square dm.
  • #1
Monoxdifly
MHB
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The volume of a cuboid box with a square base is 2 litres. The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides. Suppose the side length of its base is x and the height of the cuboid is h. The minimum production cost is reached when the surface area is ...
A. 4 square dm
B. 6 square dm
C. 8 square dm
D. 10 square dm
E. 12 square dm

What I've done:
\(\displaystyle x^2t=2→t=\frac{2}{x^2}\)
The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides. I tried to put that into account regarding the surface area f(x), and thus:
\(\displaystyle f(x)=2x^2+4(2x)t=2x^2+8x(\frac{2}{x^2})=2x^2+\frac{16}{x}\)
To determine the value of x so that the surface area will be minimum:
f'(x) = 0
\(\displaystyle 4x-\frac{16}{x^2}=0\)
\(\displaystyle 4x=\frac{16}{x^2}\)
\(\displaystyle 4x^3=16\)
\(\displaystyle x^3=4\)
\(\displaystyle x=\sqrt[3]{4}\)
Minimum surface area =\(\displaystyle 2(\sqrt[3]{4})^2+\frac{16}{\sqrt[3]{4}}=2\sqrt[3]{16}+\frac{4^2}{4^{\frac13}}=2\sqrt[3]{16}+4^{\frac53}\)

I'm stuck. Pretty sure I misinterpreted the whole "The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides." thing. What was I supposed to do?
 
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  • #2
surface area, $A = 2x^2 + 4xh$

$h = \dfrac{2}{x^2} \implies A = 2x^2 + \dfrac{8}{x}$

cost of the top and bottom is twice that of the lateral sides ...

$C = k \left(4x^2 + \dfrac{8}{x} \right)$, where $k$ is a production cost constant

$C’ = k \left(8x - \dfrac{8}{x^2} \right) = 0 \implies x = 1$
 
  • #3
Why is the 2 only multiplied by \(\displaystyle 2x^2\) and not with \(\displaystyle \frac8x\)?
 
  • #4
Monoxdifly said:
Why is the 2 only multiplied by \(\displaystyle 2x^2\) and not with \(\displaystyle \frac8x\)?

top & bottom, surface area = $2x^2$, costs twice as much to produce than the lateral sides, surface area = $4xh = 4x \cdot \dfrac{2}{x^2} = \dfrac{8}{x}$
 
  • #5
Ah. Right. That makes sense. Thank you.
 

What is minimum surface area and why is it important in science?

Minimum surface area refers to the smallest possible surface area that can enclose a given volume. It is important in science because it allows us to determine the most efficient and compact shape for a given object, which has practical applications in fields such as engineering and material science.

How is minimum surface area calculated?

The calculation of minimum surface area depends on the shape of the object. For example, for a sphere, the minimum surface area can be calculated using the formula A = 4πr², where r is the radius of the sphere. For other shapes, such as a cube or cylinder, different formulas will need to be used.

What are some real-world examples of minimum surface area?

Some real-world examples of minimum surface area include soap bubbles, which naturally form a sphere due to surface tension, and honeycombs, which have hexagonal cells that provide maximum volume with minimum surface area.

How does minimum surface area relate to the concept of optimization?

Minimum surface area is closely related to the concept of optimization, which involves finding the best solution or design for a given problem. In many cases, minimizing surface area is a key factor in optimization, as it allows for the most efficient use of materials and resources.

What are some potential challenges in achieving minimum surface area in design?

Some challenges in achieving minimum surface area in design include balancing structural integrity with minimal surface area, as well as considering practical factors such as manufacturing limitations and cost. Additionally, in some cases, the desired function or purpose of the object may conflict with achieving minimum surface area.

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