# 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

#### Answer

#### Answer:

#### Explanation

#### Explanation:

#### 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:

Let us set up the following variables:

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

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

Differentiating [1] wrt

At a critical point,

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

# :. 6x^2 = 24#

# :. x^2 = 4#

# :. x = +-2#

Obviously

# :. x = 2#

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

# :. (d^2A)/dx^2 = -12x #

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

When

Width =

#2*2 = 4#

Height =#12-4=8#

Area =#32#

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