Solution

1.

I = ⌠ ⌡ 1 0  ⌠ ⌡ x x2  x ey/x dy dx = ⌠ ⌡ 1 0  x2 (e−ex) dx = 2− 2e 3 .

2. Let

f* ≡ πr2 + a2 − λ(2 πr + 4 a − 1)

then

∂f* ∂r = 2 πr − 2 πλ = 0,     ∂f* ∂a = 2 a − 4 λ = 0,     ∂f* ∂λ = 2 πr + 4 a − 1 = 0,

from which one obtains

r = 1 2 (π+ 4) ,     a = 1 π+ 4 ,     λ = 1 2 (4 + π) .

So one can cut the string at 4 a = 4/(π+ 4)  ∼ 0.56099 m to make a square and use the rest (2 πr = π/(π+4)  ∼ 0.439901 m) to make a circle.

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File translated from T_EX by T_TH, version 4.03.
On 28 Sep 2025, 17:42.