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A fence against a wall

When a wall is one side, the fence is used only on the other three. With a fixed length of fence the width can be chosen freely, and the length is whatever remains. The area then becomes a function of one variable. You differentiate, set the derivative equal to zero and check that it is a maximum. The endpoints give zero area, so the interior point is the largest.

A(x) = x(L - 2x)

area against a wall

A'(x) = L - 4x

the derivative

Symbols

$L$	the length of fence	m
$x$	the width out from the wall	m
$A$	the area	m^2

Example

With $40$ m of fence, $A(x) = x(40 - 2x)$ . $A'(x) = 40 - 4x = 0$ gives $x = 10$ m. The length is $20$ m and the area is $200$ m $^2$ .

The wall is one long side. Then there are two widths and only one length that uses fence.

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Part of Mathematics R1: Optimisation.