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Question

The measure of the area of the square is 81-90x+25x^2 where x is a positive integer. What is the smallest possible perimeter measure for the square?

Answer 1

Area of square
= 25x^2 - 90x + 81
= (5x - 9)^2

The side of the square is 5x - 9

The smallest integral value of 5x - 9 when x is a positive integer is x = 2 ----> 5x - 9 = 1

Smallest possible side length = 1 unit.

Smallest possible perimeter = 4 units.

Answer 2

Area = s^2 = 25x^2 - 90x + 81 = (5x-9)^2 so s = 5x - 9. The smallest positive value for s where x is a positive integer is 2, which gives s = 1. The area for the square is 81 - 90 * 2 + 25 * 2^2 = 81 -180 + 100 = 1 as a check.

Therefore the perimeter = 4*s = 4.

Answer 3

25x² -90x+81 = (5x-9)²

We have been told that x is a positive integer so x ≥ 1

We want to find x ≥ 1 such that (5x -9)² is as small as possible (the smaller the area, the smaller the perimeter.

If (5x-9)² =0 that means x = 9/5 = 1.8 - not an integer - so we look at the two closest integers instead (1 and 2)

x=1 : area= (5x-9)² = 16 - sides of the square are 4x4 - so perimeter is 16
x=2 : area = (5x-9)² = 1 - sides of the square are 1x1 - so perimeter is 4

So x=2 gives the smallest solution and the smallest possible measure for the perimeter is 4

Answer 4

You have the function f(x) = 25x^2 - 90x + 81
If you take the derivative f'(x):
50x - 90
the smallest perimeter or minimum is established by putting f'(x) = 0
So, f'(x) = 0 ==> 50x - 90 = 0 ==> x = 9/5

If 25x^2 - 90x + 81 = 0 , the quadratic equation has one solution: x = 9/5.

So the perimeter would be 4x or 36/5

Answer 5

s^2 = 25x^2 -90 x + 81
s = sqr(25x^2 -90 x + 81)
4 s = 4 sqr(25x^2 -90 x + 81)