The correct answer is: [A]: " 6 to the power of (1 over 6) " .
Explanation:
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√ [(∛6)] = √ [6⁽¹/³⁾ ] = [6⁽¹/²⁾ ] ⁽¹/³⁾ = 6^ ( ⁽¹/²⁾ * ⁽¹/³⁾ ) = ???? ;
<u>Note</u>: ;
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→ 6^ ( ⁽¹/²⁾ * ⁽¹/³⁾ ) ;
= 6^ ( ⁽¹/⁶⁾ ) ;
→ which is: Answer choice: [A]: " 6 to the power of (1 over 6) " .
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Answer:
The load balance Mw minimizes the total cost
Step-by-step explanation:
<u>Optimizing With Lagrange Multipliers</u>
When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.
Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.
The cost of each generator is given by the formula
It means the cost for each generator is expanded as
The total cost of production is
Simplifying and rearranging, we have the objective function to minimize:
The restriction can be modeled as a function g(x)=0:
Or
We now construct the auxiliary function
We find all the partial derivatives of f and equate them to 0
Solving for \lambda in the three first equations, we have
Equating them, we find:
Replacing into the restriction (or the fourth derivative)
And also
The load balance Mw minimizes the total cost