Triangle Cookie Company

Let $s$ be the slope of the sides of the triangle, and let $r_3=1, r_2, r_1$ be the radii of the cookies, with the bottom row being all unit circles. The length of the base and the area of the triangle are

$b=2\left( 2+\dfrac { 1 }{ s } \left( 1+\sqrt { 1+{ s }^{ 2 } } \right) \right)$

$A=\frac { 1 }{ 2 } b\left( \dfrac { 1 }{ 2 } bs \right)$

The radii then works out to

$r_3=1$

${ r }_{ 2 }=\dfrac { \left( 2+\dfrac { 1 }{ s } \left( -1+\sqrt { 1+{ s }^{ 2 } } \right) \right) }{ \left( 1+\dfrac { 1 }{ s } \left( 1+\sqrt { 1+{ s }^{ 2 } } \right) \right) } { r }_{ 3 }$

${ r }_{ 1 }=\dfrac { \left( 1+\dfrac { 1 }{ s } \left( -1+\sqrt { 1+{ s }^{ 2 } } \right) \right) }{ \left( 0+\dfrac { 1 }{ s } \left( 1+\sqrt { 1+{ s }^{ 2 } } \right) \right) } { r }_{ 2 }$

The ratio of area of cookies to the area of the triangle then becomes

$R=\dfrac { \pi \left( 3{ { r }_{ 3 } }^{ 2 }+2{ { r }_{ 2 } }^{ 2 }+{ { r }_{ 1 } }^{ 2 } \right) }{ A }$

With $s$ as the variable, by numerical means the maximum value for $R$ can be found to be

$R=0.7201310578…$
$r_1=0.967197787…$
$r_2=0.987327667…$
$r_3=1$
$s=1.935156…$

while in the original case it was

$R=0.7199889689…$
$r_1=1$
$r_2=1$
$r_3=1$
$s=2$

so that the percent improvement is

$100 \left( \dfrac {0.7201310578…-0.7199889689… } {0.7199889689…} \right)=0.0197349...\% \; \simeq 0.02\%$