Waldo sells $x$ pieces of fencing, and each brings in revenue of $r=30-2 x$ dollars. The fixed costs are 27 dollars a week. Waldo's variable cost to make 1 fencing piece is 6 dollars.
How many pieces of fencing should Waldo make to maximize profit?
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Step 1 ：Waldo sells \(x\) pieces of fencing, and each brings in revenue of \(r=30-2 x\) dollars. The fixed costs are 27 dollars a week. Waldo's variable cost to make 1 fencing piece is 6 dollars.

Step 2 ：The profit is given by the revenue minus the costs. The revenue is given by the number of pieces sold times the price per piece, and the costs are given by the fixed costs plus the variable costs times the number of pieces made. We want to find the number of pieces that maximizes the profit.

Step 3 ：This is a problem of maximizing a function, which can be solved by finding the derivative of the function and setting it equal to zero.

Step 4 ：The profit function is \(x*(30 - 2*x) - 6*x - 27\).

Step 5 ：The derivative of the profit function is \(24 - 4*x\).

Step 6 ：We set the derivative equal to zero and solve for \(x\) to find the number of pieces that maximizes the profit.

Step 7 ：The optimal number of pieces of fencing that Waldo should make to maximize profit is 6. This is the value of \(x\) that makes the derivative of the profit function equal to zero.

Step 8 ：Final Answer: \(x=\boxed{6}\)