The wheel of Theodorus 1 inch
1 answer:
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Let 
be an arbitrary point on the surface. The distance between 
and the given point 
is given by the function
Note that
and 
attain their extrema, if they have any, at the same values of 
. This allows us to consider the modified distance function,
So now you're minimizing
subject to the constraint 
. This is a perfect candidate for applying the method of Lagrange multipliers.
The Lagrangian in this case would be
which has partial derivatives
Setting all four equation equal to 0, you find from the third equation that either
or 
. In the first case, you arrive at a possible critical point of 
. In the second, plugging into the first two equations gives
and plugging these into the last equation gives
So you have three potential points to check:, 
, and 
. Evaluating either distance function (I use 
), you find that
So the two points on the surface closest to the point are 
.
Note that
So now you're minimizing
The Lagrangian in this case would be
which has partial derivatives
Setting all four equation equal to 0, you find from the third equation that either
and plugging these into the last equation gives
So you have three potential points to check:
So the two points on the surface
Answer:
Let x be the width of the compartment,
Then according to the question,
The length of the compartment = x + 2
And the depth of the compartment = x -1
Thus, the volume of the compartment,
But, the volume of the compartment must be 8 cubic meters.
⇒
⇒
⇒
If and if
But, width can not be the complex number.
Therefore, width of the compartment = 2 meter.
Length of the compartment = 2 + 2 = 4 meter.
And, Depth of the compartment = 2 - 1 = 1 meter.
Since, the function that shows the volume of the compartment is,
When we lot the graph of that function we found,
V(x) is maximum for infinite.
But width can not infinite,
Therefore, the maximum value of V(x) will be 8.
