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A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y = 12 − x^2. What are the dimensions of such a rectangle with the greatest possible area?

1 Answer
Dec 5, 2016

Answer:

The greatest area occurs when the rectangle has a width of 4 and a height of 8 leading to a maximum area of 32

Explanation:

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Let us set up the following variables:

# {(P(x,y), "coordinate of the right hand corner"), (A, "Area of Rectangle") :} #

#P# lies on the parabola and #y=12-x^2#, so #P=P(x,12-x^2)#

Due to symmetry The width of the rectangle is half the distance between P and the y-axis, ie

Width = #2x# and Height=#y#

Hence the Area of the rectangle is:

# \ \ \ \ \ A = Wdith xx Height #
# :. A = 2xy #
# :. A = 2x(12-x^2) #
# :. A = 24x-2x^3) # ..... [1]

We are asked to maximise the Area as #x# changes so hopefully we can identify a critical point of #A# associated with a maximum, So we need to find #(dA)/dx#

Differentiating [1] wrt #x#
# :. (dA)/dx = 24-6x^2 # ..... [2]

At a critical point, # (dA)/dx = 0 #

# :. 24-6x^2 = 0#
# :. 6x^2 = 24#
# :. x^2 = 4#
# :. x = +-2#

Obviously #x# must be positive (otherwise we have an imaginary rectangle with negative area for a box that has collapsed in on itself)

# :. x = 2#

We need to check if this is a max or a min, so differentiate [2] wrt #x# to get;

# :. (d^2A)/dx^2 = -12x #
# :. (d^2A)/dx^2 = -12x < 0 " when " x=2#, confirming a max

When #x=2# we have:

Width = #2*2 = 4#
Height = #12-4=8#
Area = #32#