# Maximum and Minimum Value of Quadratic Equation

One of the most useful applications of quadratic function is its maximum and value. At the end of this post you will appreciate the importance of this feature of quadratic curve. This post will focus on algebraic approach. Using calculus this problem can be solve with less hassle. The maximum and minimum value of quadratic equation is the y-coordinate of the vertex.

Given the quadratic function $f(x)=ax^2+bx+c$, if a>0 the curve is opening upward thus the graph of the function has a minimum value. If a<0 the curve is opening downward thus the graph of the function has a maximum value.

Worked Problem 1:

Find the maximum value of $y=9-x^2$

Solution:

Using the y-coordinate of the vertex formula K

$k=\displaystyle\frac{4ac-b^2}{4a}$ , in the problem b=0.

$k=\displaystyle\frac{4ac-0^2}{4a}$

$k=\displaystyle\frac{4ac}{4a}$ 4a cancel out leaving only c.

$k=c$

$k=9$

The maximum value of the graph is 9.

Worked Problem 2:

Gorgeous April threw the ball upward with an initial velocity of 10 ft/s at a certain angle.  Amazing Cheng looked at the curve made by the motion of the ball and came up with an intuition that the height(h) in feet  of the curve from the ground  can be expressed as a function of time (t) in second with an equation $h(t)=10t-16t^2$. Assuming that his intuition is correct. What is the maximum height in feet the ball can go up from the ground?

Solution:

The equation $h(t)=10t-16t^2$ is a quadratic in t. Think t like x, and y like h(t).

The maximum height

$h=\displaystyle\frac{4ac-b^2}{4a}$, where c=0

$h=\displaystyle\frac{4a(0)-b^2}{4a}$

$h=\displaystyle\frac{-b^2}{4a}$

$h=\displaystyle\frac{-(10)^2}{4(-16)}$

$h=\displaystyle\frac{25}{16}$

The maximum height is 25/16 ft

Worked Problem 3:

John has 16 feet of fencing material to fence his rectangular garden. What is the maximum area he can fence out of this fencing material?

Solution:

Let l be the length and w be the width of the garden.

The perimeter of rectangle must be equal to the fencing material.

$P=2l+2w$

$16=2l+2w$

$8=l+w$ or $w=8-l$ eqn. 1

For the area;

$A=lw$

Substitute eqn.1 here

$A=(8-w)w$

$A=8w-w^2$

Is it familiar? Yes! This is another quadratic equation in w. Think A as y and think w as x. You can graph the equation and get the maximum value of A or we can just our formula. The vertex of the equation now is at (w,A) instead of (h,k).

Solving for maximum area:

$A=\displaystyle\frac{4ac-b^2}{4a}$, where c=0

$A=\displaystyle\frac{-b^2}{4a}$

$A=\displaystyle\frac{-(8)^2}{4(-1)}$

$A=16$

Since the given length is in feet the maximum area must be 16 ft2