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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-

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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
$$\:{Find}\:{the}\:{largest}\:{possible}\:{area} \\ $$$$\:{of}\:{trapezoid}\:{that}\:{can}\:{be}\:{drawn}\: \\ $$$$\:{under}\:{the}\:{x}−{axis}\:{so}\:{that}\:{one}\: \\ $$$$\:{of}\:{its}\:{bases}\:{is}\:{on}\:{the}\:{x}−{axis}\: \\ $$$$\:{and}\:{the}\:{other}\:{two}\:{vertices}\:{are} \\ $$$$\:{on}\:{the}\:{curve}\:{y}={x}^{\mathrm{2}} −\mathrm{9}\: \\ $$
Commented by mr W last updated on 22/Jan/23
question is wrong. with given conditions trapezoid can be ∞ large.
$${question}\:{is}\:{wrong}.\:{with}\:{given}\: \\ $$$${conditions}\:{trapezoid}\:{can}\:{be}\:\infty\:{large}. \\ $$
Commented by cortano1 last updated on 22/Jan/23
the trapezoid under x−axis
$${the}\:{trapezoid}\:{under}\:{x}−{axis} \\ $$
Commented by cortano1 last updated on 22/Jan/23
Commented by mathlove last updated on 22/Jan/23
any app used for the graph?
$${any}\:{app}\:{used}\:{for}\:{the}\:{graph}? \\ $$
Commented by mr W last updated on 22/Jan/23
but your question describes a trapezoid like this:
$${but}\:{your}\:{question}\:{describes}\:{a} \\ $$$${trapezoid}\:{like}\:{this}: \\ $$
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?
$${i}\:{wrong}\:{or}\:{you}\:{wrong}? \\ $$
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
$${i}\:{think}\:{the}\:{question}\:{required}\: \\ $$$$\:{base}\:{of}\:{trapezoid}\:{must} \\ $$$${be}\:{on}\:{curve}\:{y}={x}^{\mathrm{2}} −\mathrm{9} \\ $$
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!
$${that}'{s}\:{exactly}\:{what}\:{i}\:{showed}\:{above}: \\ $$$${one}\:{base}\:{is}\:{on}\:{the}\:{x}−{axis},\:{the} \\ $$$${other}\:{base}\:{is}\:{on}\:{the}\:{curve}. \\ $$$${bases}\:{are}\:{the}\:{two}\:{parallel}\:{sides}! \\ $$
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)))
$${get}\:\:\:−\mathrm{3}\leqslant{b}\leqslant{a}\leqslant\mathrm{3} \\ $$$${S}=\frac{\mid{a}^{\mathrm{2}} −\mathrm{9}\mid×\mid{b}^{\mathrm{2}} −\mathrm{9}\mid×\mid{a}−{b}\mid}{\mathrm{2}}=\frac{\left(\mathrm{9}−{a}^{\mathrm{2}} \right)\left(\mathrm{9}−{b}^{\mathrm{2}} \right)\left({a}−{b}\right)}{\mathrm{2}} \\ $$$$\begin{cases}{\frac{\partial{S}}{\partial{b}}=\frac{\mathrm{9}−{a}^{\mathrm{2}} }{\mathrm{2}}\left(−\mathrm{9}+\mathrm{3}{b}^{\mathrm{2}} −\mathrm{2}{ba}\right)=\mathrm{0}}\\{\frac{\partial{S}}{\partial{a}}=\frac{\mathrm{9}−{b}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{9}−\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{ab}\right)=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{b}=\:\:\:\:\mathrm{3}\:\:\:\:\:\:{or}\:\:\:\:−\sqrt{\frac{\mathrm{9}}{\mathrm{5}}}}\\{{a}=−\mathrm{3}\:\:\:\:\:{or}\:\:\:\:\:\:\:\:\:\sqrt{\frac{\mathrm{9}}{\mathrm{5}}}\:\:}\end{cases}\:\:\:\:\: \\ $$$$\Rightarrow\:\:\:\:\:\:\:{S}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{S}=\frac{\mathrm{72}}{\mathrm{25}}\sqrt{\frac{\mathrm{9}}{\mathrm{5}}} \\ $$$$\Rightarrow\Rightarrow{S}\left({max}\right)=\frac{\mathrm{72}}{\mathrm{25}}\sqrt{\frac{\mathrm{9}}{\mathrm{5}}}=\frac{\mathrm{216}}{\mathrm{25}\sqrt{\mathrm{5}}} \\ $$

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