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I let the computer do the graphing.  

f(x) = x(x + 3)²(1 - x) / 3

We see obvious zeros at x=0, x=-3 and x=1.

f(x) = x(x + 3)²(1 - x) / 3

f(x) = x(x^2 + 6x + 9)(1 - x)/3

f(x) = x(x^2 + 6x + 9 - x^3 - 6x^2 - 9x)/3

f(x) = x(- x^3 - 5x^2 -3x +9)/3

f(x) = (-x^4 - 5x^3 -3x^2 +9x)/3

f'(x) = (-4x^3 - 15x^2 -6x + 9)/3

We know there's f'(-3)=0 so x+3 must be a factor here,

f'(x) = (x + 3) (-4x^2 - 3x + 3)

We get stationary points when x = -3 or

x = (-1/8)( 3 ± √(9 - 4(-4)(3))) =  (-3 ± √57)/8  about -1.3 and 0.6 which are a local minimum and maximum respectively.

f''(x) = (-12x^2 - 30x -6)/3 = -4x^2 - 10x - 2

f''(-3) = -4(9)+ 30 - 2 = -8

Concave Down, Negative.   (-3,0) is a local maximum.

f''((-3 - √57)/8) = -4(-1.3)^2 - 10(-1.3) - 2 = 4.24

Concave up, positive, a CUP

(-3 - √57)/8, f((-3 - √57)/8) ) is a minimum.

f''((-3 + √57)/8) = -4(.6)^2 - 10(.6) - 2 = -9.44

CDN, ( ((-3 + √57)/8), f((-3 + √57)/8) ) is a maximum.

I didn't evaluate f((-3 ±√57)/8) which are around -3 and 1 but I leave that calculation to you.

69

step-by-step explanation:

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

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