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An open box with a square base must hold 4 m³. Find the least…

Step by step, with the reason for every step · Optimization · Calculus 1

Step by step

The question

Question: An open box with a square base must hold 4 m³.

Find the least material needed.

1 Draw and name

The base is a square: x by x.

The height is h.

There is no top.

x = the side of the base
h = the height

2 Write the material

The material is the base, x squared, plus 4 sides, each x by h.

S = x2 + 4xh

3 Use only one letter

S has two letters.

The volume gives the link: base times height is 4.

x2h = 4

So h = 4/x².

h = 4x2

Put it in place of h.

S = x2 + 4x × 4x2

One x cancels: S = x² + 16 over x.

= x2 + 16x

4 Which x make sense

A length is positive: x is more than 0.

x > 0 =

5 Slope = 0

The smallest value is where the slope is 0.

x² gives 2x.

16/x is 16x to the power −1, which gives −16/x².

S′(x) = 2x − 16x2
16x = 16x−1 → −16x−2

Set it to 0, and move the fraction across.

2x = 16x2

Times both sides by x², so they stay equal.

2x3 = 16

Divide by 2.

x3 = 8

The cube root of 8 is 2, because 2 × 2 × 2 = 8.

x = 3√8 = 2

6 Make sure it is the smallest

Before 2, the slope is negative: going down.

After it, positive: going up.

Down, then up: the bottom.

A minimum.

S′(1) = −14 < 0 ↘, S′(3) = 4.22 > 0 ↗

7 Answer the question

The height: 4/4 is 1.

h = 422 = 1

So the least material is 12 m²: a base 2 by 2, 1 high.

S(2) = 22 + 162 = 4 + 8 = 12 m²

8 Check

Let's check with x a bit smaller: 1 gives 17.

x = 1: S = 17

And x = 3 gives 14.33.

Both are bigger than 12.

Right!

x = 3: S = 14.33

Let's look back.

First: draw and name.

Then: write the material.

Then: use only one letter.

Then: which x make sense.

Then: slope = 0.

Then: make sure it is the smallest.

Then: answer the question.

Last: check.

Answer
S(2) = 22 + 162 = 4 + 8 = 12 m²

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