Answer :
Part A
The functions would be:
[tex]\begin{gathered} WG(x)=x+5 \\ LG(x)=3x+2 \end{gathered}[/tex]Part B
[tex]\begin{gathered} AG(x)=WG(x)\cdot LG(x) \\ \rightarrow AG(x)=(x+5)(3x+2) \\ \rightarrow AG(x)=3x^2+2x+15x+10 \\ \\ \Rightarrow AG(x)=3x^2+17x+10 \end{gathered}[/tex]Part C
Let's evaluate x = 7 in AG(x)
[tex]\begin{gathered} AG(7)=3(7^2)+17(7)+10 \\ \rightarrow AG(7)=276 \end{gathered}[/tex]Thereby, the area of the garden would be 276 square feet
Part D
The functions would be:
[tex]\begin{gathered} WB(x)=\frac{x}{2} \\ LB(x)=x+2 \end{gathered}[/tex]Part E
[tex]\begin{gathered} AB(x)=WB(x)\cdot LB(x) \\ \rightarrow AB(x)=(\frac{x}{2})(x+2) \\ \\ \Rightarrow AB(x)=\frac{x^2}{2}+x \end{gathered}[/tex]Part F
[tex]\begin{gathered} ATB(x)=x^2+AB(x) \\ \rightarrow ATB(x)=x^2+\frac{x^2}{2}+x \\ \\ \Rightarrow ATB(x)=\frac{3}{2}x^2+x \end{gathered}[/tex]Part G
[tex]\begin{gathered} AR(x)=AG(x)-ATB(x) \\ \rightarrow AR(x)=3x^2+17x+10-\frac{3}{2}x^2-x \\ \\ \Rightarrow AR(x)=\frac{3}{2}x^2+16x+10 \end{gathered}[/tex]Part H
We have a function for the area of the bell pepper patch in terms of x, the measurement of the lenght and width of the tomato patch. This is:
[tex]AB(x)=\frac{x^2}{2}+x[/tex]We know the value of this area. This way, we can solve the equation for x,
[tex]31.5=\frac{x^2}{2}+x\rightarrow63=x^2+2x\rightarrow x^2+2x-63=0[/tex]Using the cuadratic formula, and ignoring non-positive results, we'll get that
[tex]x=7[/tex]Now, plugging in this value in AR(x),
[tex]\begin{gathered} AR(7)=\frac{3}{2}(7^2)+16(7)+10 \\ \Rightarrow AR=195.5 \end{gathered}[/tex]This way, we can conclude that the remaining space in the garden after planting tomatoes and bell peppers is 195.5 square feet