FINDING MAXIMUM AND MINIMUM VALUE OF QUADRATIC FUNCTION

The vertex of a parabola is the point where the line of symmetry of the parabola intersects the parabola.

Let f be a quadratic function with standard form

y  =  a(x - h)² + k

the maximum and minimum value of f occurs at x  =  h.

In order to find the maximum or minimum value of quadratic function, we have to convert the given quadratic equation in the above form.

Minimum Value of Parabola :

If the parabola is open upward, then it will have minimum value

If a > 0, then minimum value of f is f(h)  =  k

Maximum Value of Parabola :

If the parabola is open downward, then it will have maximum value.

If a < 0, then maximum value of f is f(h)  =  k

Finding Maximum or Minimum Value of a Quadratic Function

Question 1 :

Suppose f , g, and h are defined by

f(x) = −x², g(x) = −2x², h(x) = −2x² + 1.

(a) Sketch the graphs of f , g, and h on the interval [−1, 1].

(b) Find the vertex of the graph of f , the graph of g, and the graph of h.

(d) What is the maximum value of f ? The maximum value of g? The maximum value of h?

Solution :

Graph of f(x) = −x² :

Graph of f(x) = −2x² :

Graph of f(x) = −2x² + 1 

(b)  The vertex of the parabola f(x) = −x² is at (0, 0). Hence the maximum value is 0.

The vertex of the parabola f(x) = −2x² is at (0, 0). Hence the maximum value is 0.

The vertex of the parabola f(x) = −2x² + 1 is at (0, 1). Hence the maximum value is 1.

Question 2 :

The function defined by 

f(x) = x² + 6x + 11.

(a) For what value of x does f(x) attain its minimum value?

(b) Sketch the graph.

(c)  Find the vertex of the graph of f .

Solution :

In order to convert the given quadratic function into vertex form, let us use the method completing the square method.

Let y = x² + 6x + 11

y - 11  =  x² + 6x

y - 11  =  x² + 2 ⋅ x ⋅ 3 + 3² - 3²

y - 11  =  (x + 3)² - 9

y  =  (x + 3)² - 9 + 11

y  =  (x + 3)² + 2

y = a(x - h)² + k

a =  1 > 0, the parabola is open upward, then it will have minimum value. Vertex of parabola (h, k) is (-3, 2)

When x = -3, f(x) attains its minimum value.

(b)  

Question 3 :

The height y(in feet) of a punted football is given by 

y = -16x²/2025 + 9x/5 + 1.5

Where x is the horizontal distance (in feet) from the point at which the ball is punted,

a) How high is the ball when it is punted ?

b) What is the maximum height of the punt ?

Solution :

y = -16x²/2025 + 9x/5 + 1.5

a = -16/2025, b = 9/5 and c = 1.5

a) The ball is punted at the starting position, where the horizontal distance is zero (x = 0). Substitute (x = 0) into the equation.

y = -16(0)²/2025 + 9(0)/5 + 1.5

= 1.5 feet

b)

Applying x = 114, we get

y = -102.68 + 205.2 + 1.5

= 104.02

Approximately 104.

Question 4 :

A company finds that its revenue for selling x units of one of its products is given by the quadratic function

R(x) = -0.5x² + 200x

 where R is given in dollars.

a. How many units of this product must be sold so that the maximum revenue is achieved?

b. What is the maximum revenue?

Solution :

a)

R(x) = -0.5x² + 200x

a = -0.5, b = 200

Maximum is at x = -b/2a

x = -200/2(-0.5)

= 200/1

= 200

At 200 units the maximum revenue is achieved.

b) Maximum revenue is x = 200

R(200) = -0.5(200)² + 200(200)

= -0.5(40000) + 40000

= -20000 + 40000

= 20000

Question 5 :

An electronics store sells 70 digital cameras per month at a price of $320 each. For each $20 decrease in price, about 5 more cameras per month are sold. Use the verbal model and quadratic function to determine how much the store should charge per camera to maximize monthly revenue.

Revenue (dollars) = Price (dollars/camera) ⋅ Sales (cameras)

R(x) = (320 − 20x) ⋅ (70 + 5x)

Solution :

R(x) = (320 − 20x) ⋅ (70 + 5x)

= 22400 + 1600x - 1400x - 100x²

= 22400 + 200x - 100x²

a = -100, b = 200 and c = 22400

x = -b/2a

= -200/2(-100)

= 1

When x = 1, R(1) = (320 − 20(1)) ⋅ (70 + 5(1))

= 300(70 + 5)

= 300(75)

= 22500

So, the expected maximum monthly revenue is $22500.

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