ANSWER
$173.55
Step-by-step explanation
We know that two opposite sides of the fish tank must be square. This means that the height of the tank must be the same length as one of the rectangular base's sides:
So the volume of the tank is:
![V=a\cdot a\cdot b=a^2b](https://img.qammunity.org/2023/formulas/physics/college/pfbvm90zn6r70wfz3rcfvdbue8appdumpj.png)
If glass costs $2 per square feet and metal costs $1.50 per square feet, the cost of the tank is:
![C=1.50\cdot(a\cdot b)+2\cdot(2ab+2a^2)](https://img.qammunity.org/2023/formulas/physics/college/dczlk7e3yjjd8kevsiumfujwu3gk3rbsr8.png)
We can apply distributive property and add like terms:
![\begin{gathered} C=1.50ab+4ab+4a^2 \\ C=5.50ab+4a^2 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/h29oy23jq8f0qtj5tw9kpgw3vnw1o0pac3.png)
From the volume equation we can find a relation between a and b:
![80=a^2b](https://img.qammunity.org/2023/formulas/physics/college/tbh8xiawsums0yspvj9ucgywgf0psivmrq.png)
Let's solve for b:
![b=(80)/(a^2)](https://img.qammunity.org/2023/formulas/physics/college/yw198ye0zkrkvul1a0qm0eehuhorgyga9m.png)
And replace b by this expression in the cost equation:
![C=5.50\cdot a\cdot(80)/(a^2)+4a^2](https://img.qammunity.org/2023/formulas/physics/college/h33ptu1l6p9lhzvyfex8qqkk0yku96yx0v.png)
![C=5.50\frac{80}{a^{}}+4a^2](https://img.qammunity.org/2023/formulas/physics/college/3d439xnz5mn0qcd6m0z7wmp2qaf9kky2z3.png)
Multiply 80 by 5.50:
![C=440a^(-1)+4a^2](https://img.qammunity.org/2023/formulas/physics/college/qsibr5vx5dk1aqod5ymyayktb9dgug3tn2.png)
Now, we want the minimum cost, so we have to find a minimum for this cost function in terms of a. Let's derive C:
![C^(\prime)=-440\cdot a^(-1-2)+4\cdot2\cdot a^(2-1)](https://img.qammunity.org/2023/formulas/physics/college/354twxvntnkly0kithinsa4mimyiai95ga.png)
![C^(\prime)=-440a^(-2)+8a](https://img.qammunity.org/2023/formulas/physics/college/jtohv8r5cmwi7lyr557zxky7z65ycogi9t.png)
To find the minimum we have to find for which values of a, C' is zero:
![-440a^(-2)+8a=0](https://img.qammunity.org/2023/formulas/physics/college/sqzujfe9im05s23dlix3hidslu0ka3clvb.png)
![8a=440a^(-2)](https://img.qammunity.org/2023/formulas/physics/college/275n0eeylee5fexx8wpz7ddqqvgufwn8xm.png)
Rewrite a in the denominator on the right side of the equation:
![8a=(440)/(a^2)](https://img.qammunity.org/2023/formulas/physics/college/846kstfw9ik99nrj30aen4z2sf6kwsbbj5.png)
Multiply both sides by a² and divide by 8:
![\begin{gathered} 8a^3=440 \\ a^3=(440)/(8) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/yqcbff2czxkjlat16u0aophzecjty7gcdc.png)
Take cubic root and solve:
![\begin{gathered} a^3=55 \\ a=\sqrt[3]{55} \\ a\approx3.80ft \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/hjqzwegj6sl74637kkvr1f93hfhu9fog4f.png)
If a = 3.80 ft then b is:
![b=(80)/(a^(2))=(80)/(3.8^2)\approx5.54ft](https://img.qammunity.org/2023/formulas/physics/college/vw84w56yp2soxgi37g77ykt8jty8wkotzq.png)
These side lengths would give the minimum cost for the tank. Let's find the cost:
![C=5.50ab+4a^2](https://img.qammunity.org/2023/formulas/physics/college/ohgsy4qfpqc8msl86d4pl4n0knrq0eidf9.png)
![\begin{gathered} C=5.50\cdot3.8\cdot5.54+4\cdot3.8^2 \\ C=173.546 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/hnykhpdlof1omqfsbb3dd6tuqhjteutois.png)
Rounded to the nearest cent, the minimum cost of a fish tank that has the given requirements is $173.55