# 7. Applied Maximum and Minimum Problems

by M. Bourne

The process of finding maximum or minimum values is called **optimisation**. We are trying to do things like maximise the profit in a company, or minimise the costs, or find the least amount of material to make a particular object.

These are very important in the world of industry.

### Example 1

The daily profit, *P*,
of an oil refinery is given by

P= 8x− 0.02x^{2},

where *x*
is the number of barrels of oil refined. How many barrels will give
maximum profit and what is the maximum profit?

Answer

The profit is a max (or min) if `(dP)/(dx)=0`.

`(dP)/(dx)=8-0.04x`

`=0`

when

`x=8/0.04=200`

Is it a maximum?

`(d^2 P)/(dx^2) = -0.04 < 0` * *for all* x, *so we have a maximum.

When `x = 200`, `P = $800`.

So if the company refines `200` barrels per day, the maximum profit of `$800` is reached.

Graph of `P=8x-0.02x^2`.

The maximum point, `(200, 800)` is indicated on the graph with a magenta dot.

[Go here to see another way to find the maximum or minimum value of a parabola.]

### Example 2

A rectangular storage area is to be constructed along the side of a tall building. A security fence is required along the remaining 3 sides of the area. What is the maximum area that can be enclosed with `800\ "m"` of fencing?

Answer

The area is `A = xy`

We know `2x + y = 800` so `y = 800 − 2x`

So the area is *A* = *x*(800 − 2*x*)
= 800*x* − 2*x*^{2}

To maximise the area, find when `(dA)/dx = 0`

`(dA)/(dx)=800-4x=0`

when

` x=200`

Is it a maximum?

`(d^2A)/(dx^2)=-4<0\ "for all"\ x`

So it is a maximum.

So the maximum area occurs when `x = 200`, `y = 400` and that area is:

*A* = 200 × 400 = 80 000 m^{2}
= 8 ha

Please support IntMath!

### Example 3

[This problem was presented for discussion earlier in the Differentiation introduction.]

A box
with a square base has no top. If 64 cm^{2} of material is used, what is the maximum possible volume for the
box?

Answer

The net for this box would be:

The **volume **of the box is *V *=* x*^{2}*y*

We are told that the surface area of the box is 64 cm^{2}. The area of the base of the box is *x*^{2} and the area of each side is *xy*, so the area of the base plus the area of the 4 sides is given by:

x^{2}+ 4xy= 64 cm^{2}

Solving for *y* gives:

`y=(64-x^2)/(4x)=16/x-x/4`

So the volume can be rewritten:

`V=x^2y`

`=x^2(16/x-x/4)`

`=16x-x^3/4`

Now

`(dV)/(dx)=16-(3x^2)/4`

and this is zero when

`x=+- 8/sqrt(3) ~~ 4.62`

(**Note:** The negative case has no practical meaning.)

**Is it a maximum? **

`(d^2V)/(dx^2)=-(3x)/2`

and this is negative when *x* is positive.
So it is a MAX.

So the **dimensions** of the box are:

Base 4.62 cm × 4.62 cm and sides 2.31 cm.

The **maximum possible volume** is

*V* = 4.62 × 4.62 × 2.31 ≈ 49.3 cm^{3}

*Check*: Area of material:

*x*^{2} + 4*xy* = 21.3 + 4 × 4.62 × 2.31 = 64

Checks OK.

### Search IntMath, blog and Forum

### Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Go to: Online algebra solver

### Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!