Quadratic Functions - Problems (1)

This is a tutorial on using quadratic functions to solve problems. The solutions and the explanations are detailed.

Review
the graph of a quadratic function of the form

f(x) = ax2 + bx + c

has a vertex at the point (h , k) where h and k are given by

h = -b/2a
and
k = c - b2/4a

also, k = f(h).

If a > 0, the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h = -b/2a.

If a < 0, the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h = -b/2a.

Example - Problem 1 : The profit (in thousands of dollars) of a company is given by.

P(x) = 5000 + 1000x - 5x2

where x is the amount ( in thousands of dollars) the company spends on advertising.
  1. Find the amount, x, that the company has to spend to maximize its profit.

  2. Find the maximum profit Pmax.

Solution to Problem 1:

  1. Function P that gives the profit is a quadratic function with the leading coefficient a = -5. This function (profit) has a maximum value at x = h = -b/2a
    x = h = -1000/2(-5) = 100

  2. The maximum profit Pmax, when x = 100 thousands is spent on advertising, is given by the maximum value of function P
    k = c - b2/4a

  3. The maximum profit Pmax, when x = 100 thousands is spent on advertising, is also given by P(h = 100)
    P(100) = 5000 + 1000(100) - 5(100)2 = 55000.

  4. When the company spends 100 thousands dollars on advertising, the profit is maximum and equals 55000 dollars.
  5. Shown below is the graph of P(x), notice the maximum point, vertex, at (100 , 55000).

    Graph of profit P(x).


Example - Problem 2 : An object is thrown vertically upward with an initial velocity of Vo feet/sec. Its distance S(t), in feet, above ground is given by

S(t) = -16t2 + vot.

Find vo so that the highest point the object can reach is 300 feet above ground.

Solution to Problem 2:

  1. S(t) is a quadratic function and the maximum value of S(t)is given by
    k = c - b2/4a = 0 - (vo)2 / 4(-16)

  2. This maximum value of S(t) has to be 300 feet in order for the object to reach a maximum distance above ground of 300 feet.
    - (vo)2 / 4(-16) = 300

  3. we now solve - (vo)2 / 4(-16) = 300 for vo
    vo = 64*300 = 80sqrt(3) feet/sec.

    4. The graph of S(t) for vo = 64*300 = 80sqrt(3) feet/sec is shown below.

    Graph of S(t).



More references and links on the quadratic functions in this website.