Question

Maximize the product (x+1)(y+2) given x + y = 10.

Answer 1

Solution

$\lparen x+1)\text{ \lparen y+2\rparen}$

Given that x+y = 10

make x the subject

x = 10 - y

$\begin{gathered} x+1)\text{ \lparen y+2} \\ xy+2x+y+2 \\ \lparen10-y\text{ \rparen y+2\lparen10-y\rparen+y+2} \\ 10y-y^2+20-2y+y+2 \end{gathered}$$\begin{gathered} collect\text{ like terms} \\ -y^2+10y-2y+y_+22 \\ -y^2+9y+22 \\ thus\text{ f\lparen y\rparen = 0} \\ -y^2+9y+22=0 \\ -y^2-2y+11y+22=0 \\ -y\left(y+2\right)\text{ +11\lparen y+2\rparen = 0} \\ -y+11=0,y+2=0 \\ y=11,y=-2 \end{gathered}$

Hence the maximize product will be

$\begin{gathered} x+1)\left(y+2\right) \\ since\text{ x+y=10} \\ x=10-y \\ when\text{ y = 11} \\ x=10-11=-1 \\ when\text{ y=-2} \\ x=10--2=12 \end{gathered}$$\lparen12+1)\left(-2+2\right)=13\left(0\right)=0$

when x = -1, y = 11

$\begin{gathered} \left(-1+1\text{ \rparen\lparen11+2\rparen}\right? \\ =0\left(13\right) \\ =0 \end{gathered}$

Therefore the maximize product = 0