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The area of the square is given as A = 9x^2 + 12x + 4
To minimize the area we solve the first derivative of A for x.
dA/dx = 0
=> 18x + 12 = 0
=> x = -12/18
When x = -12/18, A = 0
This proves what is evident from the fact that A represents the area, it cannot be negative. The minimum value that the area can take is 0.
The minimum area of the square is 0.
We are asked to find the minimum area of a square given that the area is expressed as A = 9x^2 + 12x +4.
We know that area cannot be negative, therefore we can make the following cases:
9x^2 + 12x + 4 > 0 or 9x^2 + 12x + 4 = 0.
Examining the two cases, we realize the minimum value would be 0 for the area of the square.
Therefore, the minimum value of the square is 0.
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