Question

A closed rectangular box has volume 28cm3. What are the lengths of the edges giving the minimum surface area? lengths =

Answer 1

For the surface of rectangle to be minimum the the dimensions must be equal. Let the length=width=height=x cm
thus the volume will be:
x×x×x=28
x³=28
x=∛28
x=2√7 cm
thus the dimensions are such that length=width=height=2√7

Answer 2

The lengths of the edges of the considered rectangular box which will give the minimum surface area are all $^3\sqrt{28}$ cms

How to find the surface area and volume of a closed rectangular box?

Suppose the dimensions of the rectangular box be 'a', 'b' and 'c' units respectively. Then, we get:

Surface area of the box = $S = 2(ab + bc + ca) \: \rm unit^2$

Volume of the box = $V = abc \: \rm unit^3$

For the considered box, let the dimensions be 'a', 'b' and 'c' cm.

Then, as we're provided that the volume of the box is 28 cubic cm, we get:
$abc = 28\\\\c = \dfrac{28}{ab} \: \rm cm$

Thus, the surface area of the considered figure would be given as:

$S = 2(ab + bc + ca) \: \rm unit^2\\\\S = 2(ab + c(b+a)) = 2(ab + \dfrac{28}{ab}(b+a))\\ S = \dfrac{2a^2b^2 + 56b + 56a}{ab} \: \rm cm^2$

There is symmetry in the equation we obtained for surface area between 'a' and 'b'.

That means, they will be of same measure to minimize or maximize the surface area. It is because if one of them is bigger than other, it is possible to prove that other one, by same tricks, can be proved bigger than first one.

Thus, we have a = b for S to be maximum or minimum.

Thus, we get:

$S_{\text{max or min}} = \dfrac{2a^4 + 112a}{a^2} = 2a^2 + 112/a = f(a)$

Taking first derivative of f(a), we get:

$f'(a) = 4a -112/a^2\\\\f''(a) = 4 +224/a^3$

Putting second rate = 0, we get critical points as:

$f'(a) = 0\\4a = 112/a^2\\a^3 = 28\\a = \: ^3\sqrt{28} = b$

At this value, we get:

$f''(\: ^3\sqrt{28}) > 0$

Thus, the only critical value found is corresponding to minima.

Thus, S is minimum when $a = b = \: ^3\sqrt{28} \: \:\rm cm$

The value of c is:

$c = \dfrac{28}{ab} = \dfrac{28}{^3\sqrt{28^2}} = (28)^{1/3} = \: ^3\sqrt{28} \: \rm cm$

Thus, the lengths of the edges of the considered rectangular box which will give the minimum surface area are all $^3\sqrt{28}$ cms

Learn more about maxima and minima of a function  here:

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