An isosceles trapezoid has one base of 12 inches and legs of 6 inches. What should its top base be in width if we wanted to maximize its area?

1 Answer
Mar 15, 2018

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Let ABCD be an isosceles trapezoid having

one base #AB=b=12"inch"#
and legs #AD=BC=c=6"inch"#

other top base #CD=a=?#
and base angles #angle ADC=angle BCD=alpha#

Its height #AE=h#

Now from figure

#h=csinalpha and DE=CF=c cosalpha#

The area of the trapezium

#A=1/2(AB+CD)xxAE#

#=>A=1/2(AB+AB+2DE)xxADsinalpha#

#=>A=1/2(12+12+2xxADcosalpha)xxADsinalpha#

#=>A=1/2(12+12+2xx6cosalpha)xx6sinalpha#

#=>A=1/2(24+12cosalpha)xx6sinalpha#

#=>A=(12+6cosalpha)xx6sinalpha#

#=>A=36(2+cosalpha)sinalpha#

#=>A=36(2sinalpha+cosalphasinalpha)#

Differentiating w r to #alpha# we get

#(dA)/(dalpha)=36(2cosalpha+cos^2alpha-sin^2alpha)#

Imposing the condition of maximization of the are #A# i.e.#(dA)/(dalpha)=0# we get

#2cosalpha+cos^2alpha-sin^2alpha=0#

#=>2cosalpha+cos^2alpha-1+cos^2alpha=0#

#=>2cos^2alpha+2cosalpha-1=0#

So #cos alpha=(-2+sqrt(2^2-4xx2(-1)))/(2xx2)#

#cos alpha <-1# not possible , so other root neglected.

#=>cos alpha=(-2+sqrt12)/(2xx2)#

#=>cos alpha=(-2+2sqrt3)/(2xx2)#

#=>cos alpha=(sqrt3-1)/2#

So #a=CD=(AB+2DE)#

#=>a=(b+2xxcxxcosalpha)#

#=>a=(12+2xx6xx(sqrt3-1)/2)#

#=>a=(12+6xxsqrt3-6))#

#=>a=6(sqrt3+1)~~16.4# inch