Question #494b1

Consider the x and y coordinates of the triangle in the first quadrant of the semi circle. The length of the required rectangle will be #2x# and the height would be #y#. [since x is measured form the centre of the circle]

So the area of the rectangle required would be #2xy#, but this area is in terms of two variables and we require it as a function of one variable for differentiation.

We know from trig and the equation of a circle that #y=rsin theta# and# x=rcos theta#. where theta is the angle made between the x axis and the radius line.

Since the radius is #8# we have #x=8costheta...........[1] and y=8sintheta............[2]# and so the area, a, of the rectangle in terms of theta is #2[8costheta8sintheta]#=#128costhetasintheta#

Let a =area, and now differentiate a with respect to theta using the product rule. #d[uv]=vdu +udv#

#da#/d#theta#= #128cos^2theta-128sin^2theta#,

#128cos^2theta=128sin^2theta# and dividing both sides by #128cos^2theta# we get
#Sin^2theta /cos^2theta =1#, [so # tan^2theta =1]#, since #tantheta# =#sin theta /costheta#.

Remember #tan^2theta =[tantheta]^2#, so by taking the square root of both sides.# tan theta =1or -1 #, but tangent is always positive in the first quadrant. #tan^_1[1]=45 degrees.#
,
so the area of the rectangle is #2xy#=#2[8cos45][8sin45]#=#128.sqrt2/2.sqrt2/2#=64. Since the angle is #45# degrees, #sin 45 =cos 45 # so #x# must equal # y # and can be found by substituting this angle in #[1]#and #[2]# above . Also since # x=y# we can say #2x^2#=#8^2# so #x=y=sqrt32#.

One could also take the second derivative to see if it is negative but is not needed here, I believe,............[it is]. Hope this helps.

We consider a rectangle inscribed in a semicircle of radius #8 \ cm#, as shown in the diagram:

Let us set up the following variables:

# { (x,"semi-width of the rectangle", cm), (y, "height of the rectangle ", cm), (A, "Area of the rectangle ", cm^2) :} #

Our aim is to find an area function, #A(x,y)# and eliminate one of the variables so that we can find the critical points wrt to the remaining variable.

The Area of the rectangle is given by:

# A = "width" xx "height" #
# \ \ \ = (2x)(y) #
# \ \ \ = 2xy # ..... [A}

Using Pythagoras we have:

# OP^2 = x^2 + y^2 #
# :. 8^2 = x^2 + y^2 #
# :. y^2 = 64 - x^2 #

We could substitute for #y# in [A] by writing #y=sqrt(8^2 - x^2)#, and substituting into [A] to get:

# A = 2xsqrt(64 - x^2) # ..... [B]

However, we can get a cleaner solution by considering #A^2#, so from [A] we have:

# A^2 = 4x^2y^2 #
# \ \ \ \ = 4x^2(64 - x^2) #
# \ \ \ \ = 256x^2-4x^4 #

Differentiating (Implicitly) wrt #x# we get:

# 2A (dA)/dx = 512x-16x^3 #

At a critical point (a minimum or a maximum) we require that the derivative, #(dA)/dx# vanish, thus we require:

# 2A xx 0 = 512x-16x^3 #
# :. 512x-16x^3 = 0#
# :. 16x(32-x^2) = 0#
# :. 32-x^2 = 0#
# :. x^2 = 32#
# :. x=+- 4 sqrt(2)#

Obviously we require that #x gt 0#, so we discard the negative solution leaving us with #x=4sqrt(2)#

And, with this value of #x# we find that

# A^2= 4x^2(64 - x^2) #
# \ \ \ \ = 4(32)(64-32) #
# \ \ \ \ = 4(32)(32) #
# \ \ \ \ = 2^2 \ (32^2) #

And so,

# A = (2)(32)=64 \ cm^2#

We need to establish that this value of #x# corresponds to a maximum. This should be intuitive, but we can validate via a graph of the result [B] (or we could perform the second derivative test):
graph{2xsqrt(64 - x^2) [-2, 10, -5, 70]}

And we can verify that a maximum when #x=4sqrt(2) ~~ 5.7# of #64# is consistent with the graph.

In addition to the great answers already posted, here is another attempt for a simpler solution. Consider the figure drawn below, depicting a rectangle ABCD drawn in a semicircle of radius 8cm,with its centre O

Length Ad of the rectangle would be #16 cos theta# and width CD would be #8 sin theta#

Area of rectangle would be #16 cos theta * 8 sin theta= 128 sin theta cos theta = 64 sin 2theta#

This would be maximum when #sin 2 theta#=1. which implies #2theta = pi/2#

This gives # theta = pi/4#

The maximum area would thus be #64 cm^2# with length of the rectangle being #16 cos theta= 8 sqrt2# and width #8sin theta= 4 sqrt2#