how could you find the measurements of the frame with a 16 inch perimeter antd the greatest possible area?

A square of 4 by 4 would give you the largest area

P=Perimeter

A=Area

a,b=sides of rectacangle

P=2a+2b

P=2(a+b) Divide with 2

P/2=a+b

a+b=P/2

a+b=16/2

a+b=8

b=8-a

A=a*b

A=a*(8-a)

A=8a-a^2

A=-a^2+8a

Function have extreme value in point where first derivative=0

First derivative:

dA/da=-2a+8

dA/da=0

-2a+8=0

8=2a Divide with 2

a=4in

b=8-a

b=8-4

b=4in

Second derivative:

d^2A/da^2=-2

When second derivative is negative fuction have maximum.

a=4in b=4 in

A(max)=4*4

A(max)=16in^2

To find the measurements of a frame with a 16-inch perimeter and the greatest possible area, you can follow these steps:

Step 1: Understand the problem
A frame consists of four sides, and the perimeter is the sum of the lengths of these sides. We need to find the measurements that add up to a total perimeter of 16 inches while maximizing the area.

Step 2: Determine the formula for perimeter & area
The formula for the perimeter of a shape depends on its type. Since the frame has four sides, we can consider it to be a rectangle. The formula for the perimeter of a rectangle is P = 2(length + width), where P denotes the perimeter, and length and width represent the measurements of the sides.

The formula for the area of a rectangle is A = length * width, where A represents the area.

Step 3: Solve for the variables
We know the perimeter is given as 16 inches (P = 16). We need to find the values of length and width that satisfy this condition while maximizing the area.

We can rearrange the perimeter formula to solve for one of the variables. Let's solve for length:
P = 2(length + width)
16 = 2(length + width)
8 = length + width
length = 8 - width

Substitute this expression for length in the area formula:
A = length * width
A = (8 - width) * width

Step 4: Maximize the area
To maximize the area, we can find the derivative of the area formula with respect to width, set it equal to zero, and solve for width. However, in this case, we don't need calculus since we have a quadratic equation.

Taking the derivative and setting it equal to zero would yield the same result. But in this scenario, we recognize that the area of a rectangle is maximized when the length and width are equal. Therefore, to maximize the area, length must be equal to width.

Step 5: Substitute equal values
Substitute width with length:
A = (8 - length) * length

Step 6: Maximize the area equation
Since the area equation is now a quadratic equation, we can simplify it to find the maximum value:
A = 8l - l^2

To find the maximum area, we want to find the vertex of this quadratic equation. The x-coordinate (length) of the vertex can be found using the formula: x = -b / (2a), where a is the coefficient of the x^2 term (in this case, -1) and b is the coefficient of the x term (in this case, 8).

Plugging the values into the formula, we get:
length = -8 / (2 * -1)
length = 4

Therefore, when the perimeter is 16 inches, the measurements of the frame with the greatest possible area are length = 4 inches and width = 4 inches.