The problem involves finding the uniform width $x$ of the picture frame so that the total area (including the picture and the frame) equals the allocated space of 15 square feet.

Given:

• The dimensions of the picture are $3.25 \, \text{ft} \times 3 \, \text{ft}$.
• The total allocated area is 15 square feet.
• The frame has a uniform width $x$.

Step-by-step solution:

1. Picture dimensions with the frame:
   The overall dimensions of the picture including the frame will be:

   • Height: $3.25 + 2x$ (frame added to top and bottom).
   • Width: $3 + 2x$ (frame added to both sides).

2. Total area equation:
   The area of the entire frame and picture is: $(3.25 + 2x)(3 + 2x) = 15$

3. Expand the equation: $(3.25 + 2x)(3 + 2x) = 9.75 + 6.5x + 6x + 4x^2$ Simplify the terms: $4x^2 + 12.5x + 9.75 = 15$

4. Set the equation to 0: Subtract 15 from both sides: $4x^2 + 12.5x + 9.75 - 15 = 0$ $4x^2 + 12.5x - 5.25 = 0$

5. Solve the quadratic equation: Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 4$, $b = 12.5$, and $c = -5.25$.

   First, calculate the discriminant: $b^2 - 4ac = 12.5^2 - 4(4)(-5.25) = 156.25 + 84 = 240.25$

   Now, calculate $x$: $x = \frac{-12.5 \pm \sqrt{240.25}}{8}$ $\sqrt{240.25} \approx 15.5$ $x = \frac{-12.5 \pm 15.5}{8}$

   So, the two solutions are: $x = \frac{-12.5 + 15.5}{8} = \frac{3}{8} = 0.375 \, \text{ft}$ $x = \frac{-12.5 - 15.5}{8} = \frac{-28}{8} = -3.5 \, (\text{not a valid solution, since width cannot be negative})$

Thus, the width of the frame is 0.375 feet.

Would you like more details or have any questions?

Related Questions:

1. How would the solution change if the picture's size were doubled?
2. What happens if the allocated area is increased to 20 square feet?
3. How does the quadratic formula work for negative discriminants?
4. Can you solve this problem graphically?
5. How would the solution change if the frame were not uniform in width?

Tip:

When solving word problems involving areas and dimensions, it's helpful to first express all quantities (like the total area) as equations before simplifying.