Question 670274
I already know that the maximum area is
when the yard is a square, but I will prove this.
Let {{{ L }}} = the length of a rectangular yard
Let {{{ W }}} = the width of this rectangular yard
given:
{{{ 2L + 2W = 40 }}} yds
( this is the definition of circumference of a rectangle )
Divide both sides by {{{ 2 }}}
(1) {{{ L + W = 20 }}}
(1) {{{ W = 20 - L }}}
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Let {{{ A }}} = the area of the yard
(2) {{{ A = L*W }}} ( also a definition )
substitute (1) into (2)
(2) {{{ A = L*( 20 - L ) }}}
(2) {{{ A = -L^2 + 20L }}}
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The rule is: when a quadratic equation has the form
{{{ ax^2 + b*x + c }}}, the max ( or min ) occurs where
the x-co-ordinate is at {{{ -b/(2a) }}}
In this problem,
{{{ a = -1 }}}
{{{ b = 20 }}}
{{{ -b/(2a) = -20/(-2) }}}
{{{ -b/(2a) = 10 }}}
{{{ L[max] = 10 }}}
and, since
(1) {{{ L + W = 20 }}}
(1) {{{ 10 + W = 20 }}}
(1) {{{ W = 10 }}}
Both the length and width are 10, so this is a
square yard
 {{{ A = 10*10 }}}
{{{ A = 100 }}} square yards
So, the maximum area is when all the sides are 10 yds
Here's the plot with Area on the vertical axis
and Length on the horizontal
{{{ graph( 500, 500, -5, 25, -10, 110,  -x^2 + 20x ) }}}