# 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?

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

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

#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 maxWhen

#x=2# we have:Width =

#2*2 = 4#

Height =#12-4=8#

Area =#32#

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To find the dimensions of the rectangle with the greatest possible area, we first need to express the area function in terms of one variable. Let the length of the rectangle be ( x ) (the base on the x-axis), then the height of the rectangle is ( y = 12 - x^2 ) (from the equation of the parabola).

Therefore, the area function, ( A(x) ), is given by ( A(x) = x \cdot (12 - x^2) ). To maximize this area, we take the derivative of ( A(x) ) with respect to ( x ), set it equal to zero, and solve for ( x ).

[ A'(x) = 12x - 3x^3 ]

Setting ( A'(x) = 0 ), we find the critical points:

[ 12x - 3x^3 = 0 ] [ x(4 - x^2) = 0 ]

This gives us ( x = 0 ) and ( x = 2 ) as critical points. To determine which critical point gives the maximum area, we can use the second derivative test or evaluate the function at critical points and endpoints.

[ A''(x) = 12 - 9x^2 ]

For ( x = 2 ), ( A''(2) = 12 - 9(2)^2 = -36 ), indicating a maximum.

For ( x = 0 ), ( A''(0) = 12 - 9(0)^2 = 12 ), indicating a minimum.

Thus, ( x = 2 ) gives the maximum area. Substituting ( x = 2 ) back into the equation for the height, we find the corresponding height:

[ y = 12 - (2)^2 = 8 ]

Therefore, the dimensions of the rectangle with the greatest possible area are: base ( x = 2 ) units, and height ( y = 8 ) units.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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