a wire whose length is at most 114cm,is bent to form a rectangle.If the length of the rectangle is 5cm longer than the width, what is the maximum area of the rectangle?

Question

a wire whose length is at most 114cm,is bent to form a rectangle.If the length of the rectangle is 5cm longer than the width, what is the maximum area of the rectangle?

Answer 1

width -- x

length --- x+5 , it said so

So, 2 widths + 2 lengths = 114

form your equation, find the width and length
Area = length x width

Answer 2

Let's solve this step-by-step:

Step 1: Let's assume the width of the rectangle as "w" cm.

Step 2: According to the problem, the length of the rectangle is 5cm longer than the width. Therefore, the length will be (w + 5) cm.

Step 3: The perimeter of the rectangle is given by the formula: 2(width + length). In this case, the perimeter will be 2(w + (w + 5)) = 2(2w + 5) = 4w + 10 cm.

Step 4: We know that the wire's length is at most 114 cm. So, the perimeter of the rectangle (4w + 10) should be less than or equal to 114 cm. Mathematically, this can be written as: 4w + 10 ≤ 114.

Step 5: Now, let's solve the inequality: 4w + 10 ≤ 114.
Subtract 10 from both sides: 4w ≤ 104
Divide both sides by 4: w ≤ 26

Step 6: From step 5, we obtained the maximum value of 'w' as 26 cm.

Step 7: Using the width value of 26 cm, we can find the length of the rectangle: w + 5 = 26 + 5 = 31 cm.

Step 8: The area of the rectangle is given by the formula: Area = length × width. Therefore, the maximum area of the rectangle is: 31 cm × 26 cm = 806 cm².

So, the maximum area of the rectangle is 806 cm² when the width is 26 cm.

Answer 3

To find the maximum area of the rectangle, we need to determine the dimensions (length and width) that will result in the largest possible area.

Let's assume that the width of the rectangle is "x" cm.

According to the given information, the length of the rectangle is 5 cm longer than the width. So, the length can be expressed as "x + 5" cm.

The perimeter of a rectangle is given by the formula: 2(length + width).

In this case, the perimeter of the rectangle is the length of the wire, which is 114 cm. So, we can set up the following equation:

2(x + 5 + x) = 114

Simplifying the equation:
2(2x + 5) = 114
4x + 10 = 114
4x = 104
x = 26

Now that we have found the width (26 cm), we can calculate the length:
Length = Width + 5 = 26 + 5 = 31 cm

Finally, we can calculate the area of the rectangle:
Area = Length * Width = 31 * 26 = 806 cm^2

Therefore, the maximum area of the rectangle is 806 square cm.