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Let's solve part (a) of the question.

The profit function of the company is given by:

\[
P(x, y) = 1200 + 80x - 2x^2 + 100y - y^2
\]

where \(x\) and \(y\) represent the cost of a unit of labor and a unit of goods, respectively. We need to maximize this profit function by finding the critical points of \(x\) and \(y\), which involves calculating the first partial derivatives of \(P(x, y)\) with respect to \(x\) and \(y\) and setting them equal to zero.

### Step 1: First partial derivative with respect to \(x\)
\[
\frac{\partial P}{\partial x} = 80 - 4x
\]

### Step 2: First partial derivative with respect to \(y\)
\[
\frac{\partial P}{\partial y} = 100 - 2y
\]

### Step 3: Set the partial derivatives equal to zero to find the critical points
\[
80 - 4x = 0 \quad \Rightarrow \quad x = \frac{80}{4} = 20
\]

\[
100 - 2y = 0 \quad \Rightarrow \quad y = \frac{100}{2} = 50
\]

Thus, the values of \(x\) and \(y\) that maximize profit are \(x = 20\) and \(y = 50\).

### Step 4: Verify that these are maximum points by checking the second derivatives (optional)

We can compute the second-order partial derivatives to ensure that these points indeed correspond to a maximum. However, the given function's structure (with negative coefficients for the quadratic terms in both \(x\) and \(y\)) suggests it is a concave function, meaning the critical points correspond to a maximum.

Would you like to see the full second derivative verification?

Here are 5 questions to expand on this:
1. What is the significance of setting the first derivative equal to zero in optimization problems?
2. How do you interpret the result of \(x = 20\) and \(y = 50\) in a real business context?
3. How would you check if the critical points correspond to a maximum, minimum, or saddle point?
4. Can this method be applied to non-quadratic profit functions, and how?
5. How would constraints on labor or goods (if given) affect this optimization problem?

**Tip**: Always check for second-order conditions when optimizing to confirm whether a critical point is a maximum or minimum.

Suppose that the profit of a certain company is given by P(x, y) = 1200 + 80x - 2x^2 + 100y - y^2. Where x and y represent the cost of unit of labor and unit of goods, respectively. Find x and y that maximize the profit. Find the maximum profit.

Solve a profit maximization problem using partial derivatives. Learn how to find critical points by taking the first partial derivatives of a quadratic profit function and setting them to zero. Suitable for undergraduate students studying calculus and optimization.

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