Can you help me on this pre calculus problem please?

Subtract the error from the value and square the resulting number, (finding the area of the sqare)

20.7375²=430.0439063

Now add the error from the value and square the resulting number.

20.8625²=435.2439063

These are the boundaries of your interval.

in general if the error in the sides is d, we have:

(20.8 + d)^2 = (20.8)^2 + 41.6d + d^2 as the upper bound for the area, and

(20.8 - d)^2 = (20.8)^2 - 41.6d + d^2 as the lower bound.

as you can see when we write it this way, it's the 41.6d term that is most important, as this is much larger than the original error, whereas the d^2 term is much smaller that the error d (only 1/256 of a square inch).

it turns out that d^2 = 0.00390625, and that the interval containing the true area is centered around (20.8)^2 + d^2 (= 432.64390625)

now 41.6d = 2.6, which in the practical world is going to matter quite a bit.

the lower bound is 432.64390625 - 2.6 = 430.04390625

the upper bound is 432.64390625 + 2.6 = 435.24390625, a surprisingly large spread of over 5 square inches.

(20.8 - 1/16)^2 <= A <= (20.8 + 1/16)^2

430.0439 <= A <= 435.2439

voila

largest possible measurement 20.8 - 1/16 = 20.8625

smallest possible measurement 20.8 + 1/16 = 20.7375

largest possible area (20.8625)^2 = 435.2439 in^2

smallest possible area (20.7375)^2 = 430.0439 in^2

430.0439 > AREA > 435.2439 (should be "or equal to" signs)

Sides of square = 20.8 ± 0.0625

20.8 - 0.0625 = 20.7375 . . . . . Area = 430.04390625

20.8 + 0.0625 = 20.8625. . . . . Area = 435.24390625

430.04390625 ≤ Area ≤ 435.24390625

s=a^2 ==> diff from both sides ds=2ada ds=20.8*2*1.6=66.56

s=(a^2-ds,a^2+ds) s=(366.08,499.2)