27 lines
1.2 KiB
Plaintext
27 lines
1.2 KiB
Plaintext
-- {
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-- "problem": "The perimeter of a rectangle is 24 inches. What is the number of square inches in the maximum possible area for this rectangle?",
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-- "level": "Level 3",
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-- "type": "Algebra",
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-- "solution": "Let one pair of parallel sides have length $x$ and the other pair of parallel sides have length $12-x$. This means that the perimeter of the rectangle is $x+x+12-x+12-x=24$ as the problem states. The area of this rectangle is $12x-x^2$. Completing the square results in $-(x-6)^2+36\\le 36$ since $(x-6)^2\\ge 0$, so the maximum area of $\\boxed{36}$ is obtained when the rectangle is a square of side length 6 inches."
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-- }
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-- Note: translating this to 2x + 2y = 24, what is xy?
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import Mathlib.Data.Real.Basic
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import Mathlib.Algebra.Group.Defs
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import Mathlib.Algebra.Ring.Defs
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import Mathlib.Tactic.Linarith.Frontend
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def valid_perimeter (x y : ℕ) : Prop :=
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2 * x + 2 * y = 24
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def area (x y : ℝ) := x * y
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theorem rewrite_y_as_x: valid_perimeter x y → y = 12 - x := by
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unfold valid_perimeter
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intro p
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have h0 : 24 = 2 * 12 := by rfl
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have h1 : 2 * x + 2 * y = 2 * (x + y) := by ring
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have h2 : 2 * (x + y) = 2 * 12 → x + y = 12 := by sorry
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have h3 : x + y = 12 → y = 12 - x := by sorry
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rw[h0, h1, h2] at p
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