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Find-the-largest-possible-area-of-trapezoid-that-can-be-drawn-under-the-x-axis-so-that-one-of-its-bases-is-on-the-x-axis-and-the-other-two-vertices-are-on-the-curve-y-x-2-9-




Question Number 185453 by cortano1 last updated on 22/Jan/23
 Find the largest possible area   of trapezoid that can be drawn    under the x−axis so that one    of its bases is on the x−axis    and the other two vertices are   on the curve y=x^2 −9
Findthelargestpossibleareaoftrapezoidthatcanbedrawnunderthexaxissothatoneofitsbasesisonthexaxisandtheothertwoverticesareonthecurvey=x29
Commented by mr W last updated on 22/Jan/23
question is wrong. with given   conditions trapezoid can be ∞ large.
questioniswrong.withgivenconditionstrapezoidcanbelarge.
Commented by cortano1 last updated on 22/Jan/23
the trapezoid under x−axis
thetrapezoidunderxaxis
Commented by cortano1 last updated on 22/Jan/23
Commented by mathlove last updated on 22/Jan/23
any app used for the graph?
anyappusedforthegraph?
Commented by mr W last updated on 22/Jan/23
but your question describes a  trapezoid like this:
butyourquestiondescribesatrapezoidlikethis:
Commented by mr W last updated on 22/Jan/23
Commented by mr W last updated on 22/Jan/23
i wrong or you wrong?
iwrongoryouwrong?
Commented by cortano1 last updated on 22/Jan/23
i think the question required    base of trapezoid must  be on curve y=x^2 −9
ithinkthequestionrequiredbaseoftrapezoidmustbeoncurvey=x29
Commented by mr W last updated on 22/Jan/23
that′s exactly what i showed above:  one base is on the x−axis, the  other base is on the curve.  bases are the two parallel sides!
thatsexactlywhatishowedabove:onebaseisonthexaxis,theotherbaseisonthecurve.basesarethetwoparallelsides!
Answered by mahdipoor last updated on 22/Jan/23
get   −3≤b≤a≤3  S=((∣a^2 −9∣×∣b^2 −9∣×∣a−b∣)/2)=(((9−a^2 )(9−b^2 )(a−b))/2)   { (((∂S/∂b)=((9−a^2 )/2)(−9+3b^2 −2ba)=0)),(((∂S/∂a)=((9−b^2 )/2)(9−3a^2 +2ab)=0)) :}  ⇒ { ((b=    3      or    −(√(9/5)))),((a=−3     or         (√(9/5))  )) :}       ⇒       S=0                 S=((72)/(25))(√(9/5))  ⇒⇒S(max)=((72)/(25))(√(9/5))=((216)/(25(√5)))
get3ba3S=a29×b29×ab2=(9a2)(9b2)(ab)2{Sb=9a22(9+3b22ba)=0Sa=9b22(93a2+2ab)=0{b=3or95a=3or95S=0S=722595⇒⇒S(max)=722595=216255

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