Answer:
A. 14x14x28
B. The maximum volume is 5488 cuibic inches
Step-by-step explanation:
The problem states that the box has square ends, so you can express volume with:
![v=x^{2} y](https://tex.z-dn.net/?f=v%3Dx%5E%7B2%7D%20y)
Using the restriction stated in the problem to get another equation you can substitute in the one above:
![4x+y=84\\\\](https://tex.z-dn.net/?f=4x%2By%3D84%5C%5C%5C%5C)
Substituting <em>y</em> whit this equation gives:
![v=x^{2} (84-4x)\\\\v=84x^{2} -4x^{3}](https://tex.z-dn.net/?f=v%3Dx%5E%7B2%7D%20%2884-4x%29%5C%5C%5C%5Cv%3D84x%5E%7B2%7D%20-4x%5E%7B3%7D)
Now find the limit of <em>x</em>:
![\frac{84x^{2}-4x^{3}}{dx}=168x-12x^{2}\\\\x=\frac{168}{12}=14](https://tex.z-dn.net/?f=%5Cfrac%7B84x%5E%7B2%7D-4x%5E%7B3%7D%7D%7Bdx%7D%3D168x-12x%5E%7B2%7D%5C%5C%5C%5Cx%3D%5Cfrac%7B168%7D%7B12%7D%3D14)
Find the length:
![y=84-4(14)=28](https://tex.z-dn.net/?f=y%3D84-4%2814%29%3D28)
You can now calculate the maximum volume:
![v=(14)^{2}(28)= 5488](https://tex.z-dn.net/?f=v%3D%2814%29%5E%7B2%7D%2828%29%3D%205488)
The answer is 5. Work backwards: first add 7 to 9, then divide by 2, then subtract 3.
9+7=16
16/2=8
8-3=5
Cross Check:
5+3=8
8*2=16
16-7=9
Final Answer: 5
Answer:
3
Step-by-step explanation:
3.5*4.5=15.75 ft^2
15.75/5.25=3 bags