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Use derivatives to solve real-world problems involving maximum and minimum values in engineering and economics.
Imagine you are an engineer designing a soda can. If you could save just 0.1 cents per can by slightly changing its dimensions while keeping the volume the same, you could save a global company millions of dollars a year. How do you find that 'perfect' shape?
In calculus, optimization is the process of finding the 'best' value—either the maximum or the minimum—of a function. This happens at critical points, where the slope of the tangent line is zero. Mathematically, if we have a function , we look for values of where or where is undefined. To confirm if a point is a maximum or minimum, we use the First Derivative Test: if changes from positive to negative, you've found a local maximum. If it changes from negative to positive, it's a local minimum. In engineering, these points represent the limits of efficiency, strength, or capacity.
A company's profit is modeled by , where is the number of units sold. Find the number of units that maximizes profit.
1. Find the derivative: . 2. Set the derivative to zero: . 3. Solve for : . 4. Verify: Since (which is negative), the graph is concave down, confirming is a maximum.
Quick Check
Why do we set the derivative to zero when looking for a maximum or minimum?
Answer
Because at the peak or valley of a smooth curve, the tangent line is horizontal, meaning its slope (the derivative) is exactly zero.
Real-world problems usually involve two equations: the objective function (what you want to maximize/minimize) and the constraint (the limit you must follow). For example, if you want to maximize the area of a garden but only have 100 meters of fencing, the area is your objective and the 100m is your constraint. The secret to solving these is to use the constraint to solve for one variable, then substitute it into the objective function. This reduces the problem to a single-variable calculus problem: .
A farmer wants to fence a rectangular area against a straight river (no fence needed on the river side). He has 1200m of fencing. What dimensions maximize the area?
1. Identify variables: Let be width and be length (parallel to river). 2. Constraint: . 3. Objective (Area): . 4. Optimize: . 5. Solve: . 6. Find : .
Quick Check
If you have two variables in your objective function, what must you find before you can differentiate?
Answer
A constraint equation that allows you to express one variable in terms of the other.
In economics and high-end manufacturing, we often minimize cost functions. These are more complex because different parts of a product might cost different amounts. For a cylindrical tank, the circular top and bottom might be made of reinforced steel (expensive), while the sides are made of standard aluminum (cheap). The goal is to find the radius and height that satisfy a specific volume while keeping the total cost at its absolute minimum. This requires careful substitution and the use of the power rule on terms like .
Design a cylindrical can with volume . The material for the top and bottom costs , and the side costs . Find the radius that minimizes cost.
1. Volume Constraint: . 2. Cost Function: . 3. Substitute : . 4. Differentiate: . 5. Solve : . 6. Result: .
If and changes from negative to positive at , what is at ?
You are maximizing the volume of a box with a fixed surface area. What is the 'objective function'?
A critical point always occurs where the derivative is zero.
Review Tomorrow
In 24 hours, try to sketch the relationship between a function's graph and its derivative's graph. Where the derivative crosses the x-axis, what happens on the original graph?
Practice Activity
Find a cylindrical object in your house (like a soup can). Measure its dimensions and calculate its volume. Then, use calculus to see if its current radius actually minimizes the surface area for that volume!