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-

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Question Number 185453 by cortano1 last updated on 22/Jan/23

$$Findthelargestpossiblearea$$$$oftrapezoidthatcanbedrawn$$$$underthex-axissothatone$$$$ofitsbasesisonthex-axis$$$$andtheothertwoverticesare$$$$onthecurvey=x^2-9$$

Commented by mr W last updated on 22/Jan/23

$$questioniswrong.withgiven$$$$conditionstrapezoidcanbe\mathrm{\infty }large.$$

Commented by cortano1 last updated on 22/Jan/23

$$thetrapezoidunderx-axis$$

Commented by cortano1 last updated on 22/Jan/23

Commented by mathlove last updated on 22/Jan/23

$$anyappusedforthegraph?$$

Commented by mr W last updated on 22/Jan/23

$$butyourquestiondescribesa$$$$trapezoidlikethis:$$

Commented by mr W last updated on 22/Jan/23

Commented by mr W last updated on 22/Jan/23

$$iwrongoryouwrong?$$

Commented by cortano1 last updated on 22/Jan/23

$$ithinkthequestionrequired$$$$baseoftrapezoidmust$$$$beoncurvey=x^2-9$$

Commented by mr W last updated on 22/Jan/23

$${that}^{\prime }sexactlywhatishowedabove:$$$$onebaseisonthex-axis,the$$$$otherbaseisonthecurve.$$$$basesarethetwoparallelsides!$$

Answered by mahdipoor last updated on 22/Jan/23

$$get-3⩽b⩽a⩽3$$$$S=\frac{\mid a^2-9\mid \times \mid b^2-9\mid \times \mid a-b\mid }{2}=\frac{(9-a^2)(9-b^2)(a-b)}{2}$$$$\{\begin{array}{l}\frac{\partial S}{\partial b}=\frac{9-a^2}{2}(-9+3b^2-2ba)=0\\ \frac{\partial S}{\partial a}=\frac{9-b^2}{2}(9-3a^2+2ab)=0\end{array}$$$$\Rightarrow \{\begin{array}{l}b=3or-\sqrt{\frac{9}{5}}\\ a=-3or\sqrt{\frac{9}{5}}\end{array}$$$$\Rightarrow S=0S=\frac{72}{25}\sqrt{\frac{9}{5}}$$$$\Rightarrow \Rightarrow S\left(max\right)=\frac{72}{25}\sqrt{\frac{9}{5}}=\frac{216}{25\sqrt{5}}$$

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