Answer: We have to find the dimension of a pool, provided the total area of it is 5000 square feet, and some clues and algebraic relation between length and width:
Diagram:
The diagram according to the information provided is as follows:
The following mathematical equation relates the total area of the pool with the dimensions of the pool:
![A=5000ft^2=2[h* l]+2[w* l]+2[w* h]\Rightarrow(1)](https://img.qammunity.org/2023/formulas/mathematics/college/pr5c4dmulgvcc4ok0poaa98l3wwbnkr5o7.png)
Plugging in the known variables in equation (1) gives a solvable equation as follows:
![\begin{gathered} A=5,000ft^2=2[20ft*(w+30)]+2[w*(w+30)]+2[w*20ft] \\ \\ \\ 5000ft^2=2(20w+600)+2(w^2+30w)+2(20w) \\ \\ 5000ft^2=40w+1200+2w^2+60w+40w \\ \\ 2w^2+140w+1200=5000 \\ \\ 2w^2+140w+1200-5000=0 \\ \\ 2w^2+140w-3800=0\Rightarrow(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/as0xvy0rkx8n8dcn1l2tgul7u5lo030pe9.png)
The solution to the quadratic equation (2) is as follows:

Therefore the dimensions are:
