An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. prove that for the minimum metal sheet its depth will be half of the width
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Answer:
Let the length, width and height of the open tank be x, x and y with respectively. Then its volume is x
2
y and total surface area is x
2
+4xy.
It is given that the tank can be hold a given quantity of water. This mean that its volume is constant. Let it be V.
V=x
2
y
The cost of the material will be least if the total surface area is least. Let's denote total surface area
S=x
2
+4xy
We have to minimize S subjects to the condition that the volume is constant.
S=x
2
+4xy
S=x
2
+
x
4V
⇒
dx
dS
=2x−
x
2
4V
⇒
dx
2
d
2
S
=2+
x
3
8V
For maximum or minimum value of S
dx
dS
=0
⇒2x−
x
2
4V
=0
⇒2x
3
=4V
⇒2x
3
=4x
2
y
⇒x=2y
Clearly
dx
2
d
2
x
=2+
x
3
8V
>0 for all x.
Here S is minimum when x=2y depth of tanh is half of width.
Step-by-step explanation:
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