# The area of a square is given as 9x^2 + 12x + 4. What is the minimum area of the square?

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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.**