S. D. V. Stephens

notes / Integrated Mathematics / chapter 9: Quadratics

9.7 Quadratic Inequalities and Applications

In progress: these notes stop partway.

A quadratic inequality is solved by finding where the parabola lies above or below the -axis. The same ideas, together with the vertex, answer maximum problems about revenue, area and height.

Quadratic inequalities

Example 9.7.1. Solve .

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X: top , bottom , sides and . So

Example 9.7.2. Solve .

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X: top , bottom , sides and . So The parabola dips below the axis between and , so it is positive outside them.

Number line with an upward parabola crossing it at -5 and 4; the rays left of -5 and right of 4 are shaded, with open circles at -5 and 4.

Example 9.7.3. Solve .

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First solve the equation: Testing each region of the number line: false to the left of , true between and , false to the right of . So

Number line with -2 and 5 marked and the regions labelled F, T, F; an upward parabola dips below the line between them, and the closed interval from -2 to 5 is shaded.

Example 9.7.4. Solve .

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: no solution.

Rearranged, the inequality is , and using the factor tree , so .

There are no real roots and the parabola opens up (it is sketched as a “U”), so it is never below .

Completing the square gives a second way to solve an inequality: it turns the quadratic into a bound on an absolute value, as the first two exercises at the end of this page do.

Applications

Revenue

Revenue: , where is the number of items sold and is the price.

Example 9.7.5. The price is . Find the maximum revenue and the price for maximum revenue.

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For instance, dollars.

Domain. We need : which is zero at and at . The parabola opens down, so it is non-negative between these: This means we do not want to produce more than items or less than items, to have positive revenue.

Max revenue. The vertex is with : so the vertex is .

Downward parabola of revenue R(x) crossing the x-axis at 0 and 300, with its vertex at (150, 7500).

The price at max revenue is dollars.

Area

Example 9.7.6. Fence for a rectangular area: find the max area available with ft of perimeter.

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A rectangle whose height is labelled x and whose width is labelled y.

Constraint: This is the limiting / constraint equation.

Area: is the maximizing equation. Substituting the constraint, The vertex is at where

Thus the vertex is : the maximum area is ft² at .

Downward parabola of area A(x)=1500x-x^2 crossing the x-axis at 0 and 1500, with its vertex at (750, 562,500).

Height

Example 9.7.7. .

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Setting gives where it lands. The coefficient on gives the max height, at the vertex

Exercises

Exercise 9.7.8. Find all such that .

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. Completing the square, , so Taking square roots, , so : (Equivalently, gives : and .)

Exercise 9.7.9. Find all such that .

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so .

Exercise 9.7.10.

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Price (value at a given : ); tickets per game; each one-dollar increase loses 100 tickets, so the tickets sold are which is at . At the revenue is . We want