I Want More Pizza!

Let the angle extended by the sector be $\theta$ and the radius of the pizza be $r = \dfrac{d}{2}$ . Then, we have:

• The area of the sector $A = \dfrac {\theta r^2}{2}$
• The perimeter $18 = 2r + \theta r \quad \Rightarrow \theta = \dfrac{18}{r} - 2$

$\begin{aligned} \Rightarrow A & = \left(\frac{18}{r} - 2 \right)\frac{r^2}{2} = 9r - r^2 \quad \Rightarrow \frac {dA}{dr} = 9 - 2r \end{aligned}$

For maximum $A: \quad \Rightarrow \dfrac{dA}{dr} = 0 \quad \Rightarrow r = \dfrac {9}{2} \quad \Rightarrow d = 9 \\ \quad \quad \Rightarrow A = 9\left(\dfrac{9}{2} \right) - \left(\dfrac{9}{2} \right)^2 = \dfrac{81}{2} - \dfrac{81}{4} = \dfrac{81}{4} \\ \quad \quad \Rightarrow d + A = 9 + \dfrac{81}{4} = 9 + 20.25 = \boxed{29.25}$

Perimiter of sector=r@+2r=18, where r=radius of circle, @=theta/angle in radians.

Hence we can say that @=(18/r)-2 Area of sector=(1/2) (r^2) @. Substituting @ as given above, we get (after some simple algebra):

Area of sector=A=9r-(r^2)

Hence, the derivative= dA/dr= 9-2r. Letting dA/dr=0 gives us r=4.5.

Hence, diameter d=9 cm and A (after substituting 4.5 as r) we get as (4.5)^2=20.25 cm^2.

Hence, we get d + A = 20.25 + 9 = 29.25. ANSWER= 29.25.