Mounting a Photograph: A Quadratic Border, a Frame Clearance and a Doubling Claim
A photographer is mounting a 30 cm × 20 cm print on a rectangular mount board, leaving a uniform border of width x cm on all four sides. The studio has decided that the finished board must have an area of 1200 cm². The frame already ordered has an inner opening of exactly 40 cm × 30 cm, and the carpenter has warned that a board touching the frame on all four sides will warp, so at least 0.3 cm clearance is needed on every side. The studio assistant sets up (30 + 2x)(20 + 2x) = 1200, solves it and writes "x = 5 or x = −30, so the border is 5 cm", and orders the board cut. The accountant, who prices board by area, uses the same expression A(x) = (30 + 2x)(20 + 2x) and argues that since the border would be doubled from 5 cm to 10 cm, the board area would also double, making a wider border unaffordable. The photographer doubts both conclusions but cannot pin down exactly where the reasoning fails.
1. Solve the equation correctly, then decide whether x = 5 satisfies every condition in the problem.
Read the worked answer
Expand: 600 + 60x + 40x + 4x² = 1200, i.e. 4x² + 100x − 600 = 0, or x² + 25x − 150 = 0.
Discriminant = 625 + 600 = 1225 = 35², so x = (−25 ± 35)/2, giving x = 5 or x = −30. A border width cannot be negative, so x = −30 is rejected on geometric grounds, not because the algebra fails.
Now test x = 5 against the frame. Board sides: 30 + 10 = 40 cm and 20 + 10 = 30 cm - exactly the frame opening. Clearance is 0 cm, below the required 0.3 cm, so x = 5 must be rejected as well.
The admissible range comes from the clearance: 30 + 2x ≤ 39.4 gives x ≤ 4.7, and 20 + 2x ≤ 29.4 also gives x ≤ 4.7. The largest usable border is 4.7 cm, giving area 39.4 × 29.4 = 1158.36 cm². So the target of 1200 cm² is unattainable with this frame; the assistant solved the equation but ignored a constraint the equation did not contain.
2. Is the accountant right that doubling the border from 5 cm to 10 cm doubles the board area? Test the claim using the expression A(x).
Read the worked answer
Compute directly: A(5) = 40 × 30 = 1200 cm², while A(10) = (30 + 20)(20 + 20) = 50 × 40 = 2000 cm². The area rises by 800 cm², not by 1200 cm² - the ratio is 2000/1200 = 5/3, not 2.
Algebraically, A(2x) = (30 + 4x)(20 + 4x) = 16x² + 200x + 600, whereas 2A(x) = 8x² + 200x + 1200. These differ by 8x² − 600, which is zero only when x² = 75. So doubling the input does not double the output; the claim would hold only for a purely quadratic expression with no constant term.
The reason is structural: A(x) is a quadratic, not a proportional relation. Doubling x doubles the linear part of the growth but quadruples the x² part, while the 600 cm² of the print itself does not change at all. The accountant's error is treating a quadratic as if it were direct proportion.
3. The photographer finds a frame with an inner opening of 42 cm × 32 cm. With the border at x = 5 cm, does the mounting now satisfy the carpenter's rule? Give the clearance and the board dimensions.
Read the worked answer
With x = 5 the board is 40 cm × 30 cm, area 1200 cm² as originally planned, so the quadratic condition is met (x = 5 is a genuine root of x² + 25x − 150 = 0).
Clearance = (42 − 40)/2 = 1 cm horizontally and (32 − 30)/2 = 1 cm vertically. Since 1 cm ≥ 0.3 cm on every side, the carpenter's warping rule is satisfied with room to spare.
The judgement is that the equation was never the difficulty - its roots were correct. The difficulty was that the frame imposed an inequality, 30 + 2x ≤ 39.4 and 20 + 2x ≤ 29.4, which the old frame could not satisfy simultaneously with the 1200 cm² target. Changing the frame loosens the inequality, and x = 5 becomes admissible. A board cut to 40 × 30 is the correct order.