AMC 8 · 2008 · #17
Easy mode Grade 4Problem
Ms. Osborne asks each student to draw a rectangle.
The rectangle has to follow two rules. Its side lengths must be whole numbers. Its perimeter must be exactly 50 units.
Each student then finds the area of the rectangle they drew. Different students can draw different rectangles, so the areas can be different too.
What is the difference between the largest area and the smallest area that are possible?
Pick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Each student draws a rectangle with whole-number side lengths and perimeter $50$. They each compute the area. What is the largest possible area minus the smallest possible area?
Givens: The rectangle has integer side lengths; The perimeter is $50$ units; Answer choices: (A) $76$, (B) $120$, (C) $128$, (D) $132$, (E) $136$
Unknowns: The difference between the largest and smallest possible areas
Understand
Restated: Each student draws a rectangle with whole-number side lengths and perimeter $50$. They each compute the area. What is the largest possible area minus the smallest possible area?
Givens: The rectangle has integer side lengths; The perimeter is $50$ units; Answer choices: (A) $76$, (B) $120$, (C) $128$, (D) $132$, (E) $136$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #2 Make a Systematic List
The perimeter pins down $l + w = 25$. The area $l \times w$ then depends on how we split $25$ into two positive integers. Tool #14 (Extreme Principle) is the right tool because we want the max and min of a product whose two factors have a fixed sum: the product is biggest when the factors are as close as possible, and smallest when they are as far apart as possible. Tool #4 (Introduce a Variable) lets us name the sides $l$ and $w$, and Tool #2 (Make a Systematic List) confirms the extremes by checking the endpoint pairs.
Execute — Answer: D
3.MD.D.8 Step 1 - Name the sides and turn the perimeter into a simple sum.
- Let $l$ and $w$ be the length and width.
- The perimeter is $2(l+w)=50$, so $l+w=25$.
- Both must be positive integers.
💡 Using $P = 2(l+w)$ for a rectangle is the Grade 3 perimeter standard. Dividing by $2$ turns the constraint into a single sum.
3.MD.C.7 Step 2 - Use the extreme principle on a product with a fixed sum.
- With $l + w = 25$ fixed, write $w = 25 - l$, so the area is $A = l(25-l)$.
- As $l$ moves from the middle ($12$ or $13$) toward the ends ($1$ or $24$), the two factors spread apart and the product shrinks.
- The biggest product happens when the factors are as close as possible.
💡 Area of a rectangle is length $\times$ width (Grade 3). Among integer splits of $25$, the closest pair is $(12,13)$ and the farthest pair is $(1,24)$.
3.MD.C.7 Step 3 - List the endpoint cases and compute.
- The pair $(12,13)$ gives the biggest area; the pair $(1,24)$ gives the smallest.
- Quickly check the list of integer pairs to confirm these are the extremes.
💡 Listing the pairs makes the pattern visible: products grow $24, 46, 66, \dots$ as the gap shrinks, peaking at $156$ for $(12,13)$.
4.OA.A.3 Step 4 Subtract to get the difference asked for.
💡 The multi-step word problem ends in one subtraction (Grade 4): max area minus min area.
3.MD.D.8 Name the sides and turn the perimeter into a simple sum. Let $l$ and $w$ be the 3.MD.C.7 Use the extreme principle on a product with a fixed sum. With $l + w = 25$ fixed 3.MD.C.7 List the endpoint cases and compute. The pair $(12,13)$ gives the biggest area; 4.OA.A.3 Subtract to get the difference asked for. Review
Reasonableness: The answer $132$ matches choice (D). Spot-check the trend with the middle of the list: $(10,15)$ gives $150$, $(11,14)$ gives $154$, $(12,13)$ gives $156$ — the product climbs as the two sides get closer, then there is no $l=w$ case since $25$ is odd. On the other end, $(1,24)=24$ and $(2,23)=46$ confirm the smallest area is at the most lopsided pair. So $156-24=132$ is consistent.
Alternative: Tool #3 (Set Up an Equation): substitute $w = 25 - l$ into $A = lw$ to get $A(l) = 25l - l^2 = -(l - 12.5)^2 + 156.25$. This parabola opens downward with peak at $l=12.5$, so the integer maximum is at $l=12$ or $13$ (both give $156$). The integer minimum on $1 \le l \le 24$ is at the endpoints $l=1$ or $24$ (both give $24$). Difference is again $156-24=132$.
CCSS standards used (min grade 4)
3.MD.D.8Solve real world and mathematical problems involving perimeters of polygons (Using $P = 2(l+w) = 50$ to reduce the constraint to $l + w = 25$ with positive integer sides.)3.MD.C.7Relate area to the operations of multiplication and addition (Writing the area as $A = l \times w$ and computing the products $12 \times 13 = 156$ and $1 \times 24 = 24$.)4.OA.A.3Solve multistep word problems posed with whole numbers using the four operations (Combining the perimeter setup, the extreme cases, and the final subtraction $156 - 24 = 132$ into one multi-step word problem.)
⭐ When two whole numbers add to a fixed total, their product is biggest when they are as close as possible and smallest when they are as far apart as possible.
⭐ When two whole numbers add to a fixed total, their product is biggest when they are as close as possible and smallest when they are as far apart as possible.
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