# Tutor profile: Cameron C.

## Questions

### Subject: Industrial Engineering

You are studying your company's product line to optimize profit. You currently produce two products using capabilities at three plants. The first product requires the use of plant 1 (4 hour/item) and plant 3 (1 hours/item). The second product requires the use of plant 2 (5 hours/item) and plant 3 (2 hours/item). We can adjust the production of either product, but are constrained by the capacity of the plants. Plant 1 has 10 hours/week available, Plant 2 has 16 hours/week available, and Plant 3 has 21 hours available. Knowing that each product 1 item provides $2,500 profit and each product 2 item provides $6,000 profit, formulate a linear programming problem to maximize profit.

The first step is to define the decision variables. Let x_{i} = number of items of product i produced per week (in thousands). Next we can define the objective function. Since we want to maximize profits, we can write it asL Max Z = 2.5x_{1} + 6x_{2} Finally, we must define the constraints based in the plant capacity: 4x_{1} \leq 10 (plant 1 constraint) 5x_{2} \leq 16 (plant 1 constraint) x_{1} + 2x_{2} \leq 21 (plant 1 constraint) Additionally, the production of each product cannot be negative, so we also set: x_{1} \geq 0 and x_{2} \geq 0

### Subject: Calculus

Given the curve y = 2x^{4} + 3x^{2} - 18, calculate the point(s) at which the tangent line is horizontal.

The derivative of a function can be interpreted as the slope of a tangent. We also know that the slope of a horizontal line is zero. Therefore, we first need to calculate the derivative of the curve and then determine at which points the derivative equals zero. The derivative of y = 2x^{4} + 3x^{2} - 18 is y' = 8x^{3} + 6x. This can equivalently be written as y' = 2x(4x^{2} + 3). From this notation we can see that y' = 0 when 2x = 0 or if 4x^{2} + 3 = 0. Solving for x, we find x = 0 and x = \pm \frac{\sqrt{3}}{2}. Therefore, the points at which the tangent line is horizontal are (0, -18), (\frac{\sqrt{3}}{2}, -19.125), and (-\frac{\sqrt{3}}{2}, -19.125)

### Subject: Statistics

Assume there are 49 students in your high school senior class. The average weight of these students is 120 pounds with a standard deviation of 30 pounds. Construct a 95% confidence interval estimate of the population mean weight of high school seniors based on this sample.

The construction of a confidence interval for the population mean is calculated as: \bar{x} - E < \mu < \bar{x} + E where E = $$z_{\alpha /2} \cdot \frac{\sigma }{\sqrt{n}} We know that n = 49, \bar{x} = 120, \sigma = 30, and \alpha = 0.05 Using a Standard Normal Distribution Table, we can lookup the value of $$z_{\alpha /2} as 1.96 Using this information, we can calculate E: E = 1.96 \cdot \frac{15}{\sqrt{576}} = 8.4 Therefore, the 95% confidence interval estimate of the population mean weight would be (111.6, 128.4)

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