Respuesta :
Answer:
The dimensions of the rectangular tank is 30.24 ft by 30.24 ft by 7.56 ft.
Step-by-step explanation:
Given that,
A rectangular tank is to be constructed with a square base and a volume of 6912 ft³.
Let length of the one side of the base be x and the height of the tank be y.
Then, the volume of the tank is= area of the base × height
=(x²)×y
=x²y ft²
Then,
x²y = 6912
[tex]\Rightarrow y=\frac{6912}{x^2}[/tex]
The surface area of the tank is
= surface area of the sides + surface area of the base
=4(xy)+x² [ surface area of each wall= length × width =xy]
=4xy+x²
A= 4xy + x²
Plug [tex]y=\frac{6912}{x^2}[/tex] in the above expression
[tex]A=4x\times \frac{6912}{x^2}+x^2[/tex]
[tex]\Rightarrow A=\frac{27648}{x}+x^2[/tex]
Differentiating with respect to x
[tex]A'=-\frac{27,648}{x^2}+x[/tex]
Again differentiating with respect to x
[tex]A''=-\frac{(-2)27648}{x^{2+1}}+1[/tex]
[tex]\Rightarrow A''=\frac{55,296}{x^3}+1[/tex]
For maximum or minimum, A'=0
[tex]-\frac{27,648}{x^2}+x=0[/tex]
[tex]\Rightarrow x=\frac{27,648}{x^2}[/tex]
[tex]\Rightarrow x^3={27,648}[/tex]
[tex]\Rightarrow x=\sqrt[3]{27,648}[/tex]
⇒x ≈ 30.24
Now
[tex]\Rightarrow A''|_{x=30.24}=\frac{55,296}{(30.24)^3}+1>0[/tex]
So, at x=30.24 ft the surface area minimum.
The length of one side of the base of the tank is = 30.24 ft
Then the height of the rectangular tank is
[tex]y=\frac{6912}{x^2}[/tex]
[tex]=\frac{6912}{30.24^2}[/tex]
=7.56 ft
The dimensions of the rectangular tank is 30.24 ft by 30.24 ft by 7.56 ft.
