The area of a particular parallelogram is given by the formula A=x^2−bx+c.
How can you rearrange this formula to correctly find the x-value that will yield the maximum area?
To rearrange the formula A=x^2−bx+c in order to find the x-value that will yield the maximum area, we will need to complete the square. The general steps are as follows:
Step 1: Rewrite the formula in the form A = a(x - h)^2 + k
Step 2: Identify the values of a, h, and k
Step 3: Solve for the x-value that will maximize the area
Let's go through each step in detail:
Step 1: Rewrite the formula in the form A = a(x - h)^2 + k
To complete the square, we want to express the quadratic equation in the form A = a(x - h)^2 + k, where a, h, and k are constants.
Given A = x^2 - bx + c, we can rewrite it as:
A = (x^2 - bx) + c
Step 2: Identify the values of a, h, and k
Comparing this to the form A = a(x - h)^2 + k, we can see that a = 1.
Next, we need to find the values of h and k. We can do this by halving the coefficient of x (b) and squaring it:
h = (b/2)^2
k = c - (b/2)^2
Step 3: Solve for the x-value that will maximize the area
If the formula is in the form A = a(x - h)^2 + k, then the maximum or minimum value occurs at x = h.
In our case, x = h will give us the x-value that yields the maximum area.
So, the rearranged formula is:
A = (x - (b/2))^2 + (c - (b/2)^2)
To find the x-value that will maximize the area, we need to solve the equation x - (b/2) = 0:
x = (b/2)
Thus, the x-value that will yield the maximum area is (b/2).
A=(x−b/2)^2+4c−b/4
first of all, x^2 - bx + c is a parabola opening upwards, so it doesn't have a
maximum, it has a minimum
dA/dx = 2x - b
= 0 for a min of A
2x - b= 0
2x= b
x = b/2