Answer :
To solve this problem you must apply the proccedure shown below:
1. You must use the formula for calculate the surface area and the volume of the tank:
[tex] 1) SA=x^{2} +4xy\\ 2) V=x^{2} y\\ 64=x^{2} y [/tex]
2. Solve for [tex] y [/tex] in the second equation and substitute it into the first one.Then, you have:
[tex] y=\frac{64}{x^{2}} \\ SA=x^{2} +4x(\frac{64}{x^{2}}) [/tex]
3. Now, derivate:
[tex] SA'=2x-256x^{-2} \\ 2x-256x^{-2}=0
[/tex]
4. When you solve for [tex] x [/tex], you obtain:
[tex] 2x^{3}-256=0\\ x=\sqrt[3]{\frac{256}{2}} \\ x=5.03 [/tex]
5. Now, you must calculate [tex] y [/tex]:
[tex] y=\frac{64}{(5.03)^{2}} \\ y=2.52 [/tex]
Therefore, the dimensions are:
The sides of the base of the tank: [tex] 5.03 feet [/tex]
The height of the tank: [tex] 2.52 feet [/tex]
Answer:
The sides of the base of the tank: 5.03 feet
The height of the tank: 2.52 feet
Step-by-step explanation:
Given:
Volume of an open-top, square-based, rectangular metal tank = 64 [tex]ft^{3}[/tex]
Let:
The side of the square base = x ft
Height of the tank = y
So, The volume of the tank = [tex]x^{2} y[/tex]
[tex]64=x^{2} y[/tex] ........(i)
And surface area = [tex]x^{2} +4xy[/tex]
From (i)
Surface area (SA)= [tex]x^{2} +4x(\frac{64}{x^{2} } )[/tex]
Now, Differentiate
(SA)' = [tex]2x-256x^{-2}[/tex]
[tex]0=2x-256x^{-2}[/tex]
solve for , you obtain:
[tex]2x^{3} -256=0\\x=\sqrt[3]{\frac{256}{2} } \\x=5.03[/tex]
Now, you must calculate
[tex]y=\frac{64}{x^{2} } \\y=\frac{64}{5.03^{2} } \\y=2.52[/tex]
Therefore, the dimensions are:
The sides of the base of the tank: 5.03 feet
The height of the tank: 2.52 feet
For more information:
https://brainly.com/question/2273504?referrer=searchResults