Cho hình chóp S.ABCD có đáy ABCD là hình thang, AD song song với BC, AD=2BC. Gọi E, F là hai điểm lần lượt nằm trên các cạnh AB và AD sao cho

Đặt \[\frac{{AE}}{{AB}} = x,\,\,\frac{{AF}}{{AD}} = y\,\,(0 < x,\,\,y \le 1)\]Theo bài ra ta có\[\frac{{3AB}}{{AE}} + \frac{{AD}}{{AF}} = 5\]

\[ \Rightarrow \frac{3}{x} + \frac{1}{y} = 5\,\,\,\left( 1 \right)\]

Vì hai khối chóp S.BCDFE và S.ABCD có cùng chiều cao nên

\[k = \frac{{{V_{S.BCDFE}}}}{{{V_{S.ABCD}}}} = \frac{{{S_{BCDFE}}}}{{{S_{ABCD}}}}\]

Đặt\[{S_{ABCD}} = S\] kẻ\[BH \bot AD\,\,\left( {H \in AD} \right)\]ta có

\[S = \frac{1}{2}BH.\left( {BC + AD} \right) = \frac{3}{2}.BH.BC\]

Ta có:\[\frac{{{S_{AEF}}}}{{{S_{ABD}}}} = \frac{{\frac{1}{2}AE.AF.\sin \angle BAD}}{{\frac{1}{2}AB.AD.\sin \angle BAD}} = xy \Rightarrow {S_{AEF}} = xy.{S_{ABD}}\]

Mà\[{S_{ABD}} = \frac{1}{2}BH.AD\]nên

\[{S_{AEF}} = \frac{1}{2}xy.BH.AD = xy.BH.BC = \frac{3}{2}BH.BC.\frac{2}{3}xy \Rightarrow {S_{AEF}} = \frac{2}{3}xy.S\]

\[ \Rightarrow {S_{BCDFE}} = {S_{ABCD}} - {S_{AEF}} = S - \frac{2}{3}xy.S = S\left( {1 - \frac{2}{3}xy} \right)\]

\[ \Rightarrow k = \frac{{S.\left( {1 - \frac{2}{3}xy} \right)}}{S} = 1 - \frac{2}{3}xy\]

Theo (1) ta có:\[\frac{3}{x} + \frac{1}{y} = 5 \Leftrightarrow y = \frac{x}{{5x - 3}}\]

Ta có

\[0 < \frac{x}{{5x - 3}} \le 1 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{x}{{5x - 3}} > 0}\\{\frac{{x - 5x + 3}}{{5x - 3}} \le 0}\end{array}} \right.\]</>

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{5x - 3 > 0\,(do\,\,x > 0)}\\{3 - 4x \le 0}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x > \frac{3}{5}}\\{x \ge \frac{3}{4}}\end{array}} \right. \Leftrightarrow x \ge \frac{3}{4}\)

Khi đó ta có

\[\begin{array}{*{20}{l}}{k = 1 - \frac{2}{3}xy = 1 - \frac{2}{3}x.\frac{x}{{5x - 3}}}\\{\,\,\,\, = 1 - \frac{{2{x^2}}}{{3\left( {5x - 3} \right)}} = \frac{{15x - 9 - 2{x^2}}}{{3\left( {5x - 3} \right)}} = f\left( x \right)}\end{array}\]

Xét hàm số\[f\left( x \right) = \frac{{ - 2{x^2} + 15x - 9}}{{3\left( {5x - 3} \right)}}\]với \[\frac{3}{4} \le x \le 1\]ta có:

\[f\prime (x) = {\frac{{( - 4x + 15).3(5x - 3) - ( - 2{x^2} + 15x - 9).15}}{{9{{(5x - 3)}^2}}}^{}}\]

\[f\prime (x) = \frac{{3( - 20{x^2} + 87x - 45) - ( - 30{x^2} + 225x - 135)}}{{9{{(5x - 3)}^2}}}\]

\[f\prime (x) = \frac{{ - 30{x^2} + 36x}}{{9{{(5x - 3)}^2}}} = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{6}{5}\left( {ktm} \right)}\\{x = 0\left( {ktm} \right)}\end{array}} \right.\]

BBT:

\[ \Rightarrow {k_{\min }} = \frac{1}{2},\,\,{k_{\max }} = \frac{2}{3}\]

Vậy\[{k_{\min }} + {k_{\max }} = \frac{1}{2} + \frac{2}{3} = \frac{7}{6}\]