The area of Sarah's garden is given by A = 3(4 - x^2). Find the maximum area of her garden.?

Method 1 : Algebraic method

A = 3 (4 - x²)
A = 12 - 3x²

For all real values of x, 3x² ≥ 0
Hence, A = 12 - 3x² ≤ 12

The maximum area of Sarah's garden = 12 (square units)


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Method 2 : Differentiation

A = 3 (4 - x²)
A' = 3 (-2x) = -6x
A" = -6

When x = 0: A' = 0 and A" < 0
Hence, maximum A when x = 0
Maximum A = 3 (4 - 0²) = 12 (square units)

A = 3(4 - x²)
For max area find derivative of A and equate to 0
dA = 3(-2x)
dA = -6x
0 = -6x
x=0

For max area, A = 3(4-0²) = 12 sq units

This can be handled in a simple algebraic way. The vertex of the parabola y = ax2 + bx + c occurs when x = -b/(2a). In this problem b = 0, so the maximum occurs when x = 0 and A = 12

A = 3(4 - x²) = 12 – 3x²
differentiate and set equal to zero to get max/min
dA/dx = –6x = 0
x = 0
in other words, there is a max, and at x = 0
which is A = 12

you can see that by looking at the equation, A = 12 – 3x²
for x = 0, the A = 12, and as x increases, the area decreases, hitting zero for x = 2.

for x values <0 or >2, the function is not defined, as area cannot be negative.

Her area grows with x until the rate of increase dA/dx = 0 and it grows no more. That's the max.

We have A = 12 - 3x^2

So we have dA/dx = 0 = - 6x and x = 0.

There is no max on x. In fact, when we look at the A function we see that it forms a parabola, which has no maxium.