Armon specializes in framed art. For a specific design of framed art, the width of the frame is 10 inches more than the width of the art inside the frame, and the length of the frame is twice as long as the frame's width. If x represents the width of the art inside the frame, what function could be used to model the area of the entire design, frame and art included?

1 answer

To model the area of the entire design, including both the framed art and the frame, we need to express the dimensions of the frame in terms of \( x \), the width of the art inside the frame.

  1. Width of the Frame:

    • Since the width of the frame is 10 inches more than the width of the art inside the frame, we can express the width of the frame as: \[ \text{Width of the frame} = x + 10 \]
  2. Length of the Frame:

    • The length of the frame is given as twice the width of the frame. Therefore, the length of the frame can be calculated as: \[ \text{Length of the frame} = 2 \cdot (x + 10) = 2x + 20 \]
  3. Dimensions of the Entire Design:

    • The dimensions of the entire design include both the art and the frame around it.
    • The total width of the framed artwork is the width of the art plus the frame on either side (the frame adds 10 inches on each side): \[ \text{Total Width} = x + 10 + 10 = x + 20 \]
    • The total length of the framed artwork is the length of the art plus the frame on the top and bottom (the frame adds the double length of the frame): \[ \text{Total Length} = (2x) + (2 \cdot 10) = 2x + 20 \]
  4. Area of the Entire Design:

    • The area \( A \) of the entire design (art and frame) can therefore be expressed as: \[ A = \text{Total Width} \times \text{Total Length} = (x + 20)(2x + 20) \]
  5. Expanded Area Function:

    • We can expand this expression: \[ A = (x + 20)(2x + 20) = x(2x + 20) + 20(2x + 20) \] \[ = 2x^2 + 20x + 40x + 400 \] \[ = 2x^2 + 60x + 400 \]

Thus, the function that models the area of the entire design, frame and art included, is: \[ A(x) = 2x^2 + 60x + 400 \]