جواب تمرین صفحه 120 درس 5 ریاضی دوازدهم تجربی (کاربرد مشتق)

الف 

\(\begin{array}{l}x\;y = 10,000 \Rightarrow y = \frac{{10,000}}{x}\\\\P(x) = 2(2,000,000x) + 2(8,000,000 \times \frac{{10,000}}{x})\\\\ = \frac{{4 \times {{10}^6}\;({x^2} + 40,000)}}{x}\end{array}\)

ب 

\(\begin{array}{l}P'(x) = 4 \times {10^6}(\frac{{2{x^2} - {x^2} - 40,000}}{{{x^2}}}) = 4 \times {10^6}(\frac{{{x^2} - 40,000}}{{{x^2}}})\\\\\mathop \Rightarrow \limits^{P'(x) = 0} \;\;4 \times {10^6}(\frac{{{x^2} - 40,000}}{{{x^2}}}) = 0\\\\{x^2} = 40,000 \Rightarrow x = 200 \Rightarrow y = 50\end{array}\)

الف 

\(\begin{array}{l}{h^2} + {x^2} = {50^2} \Rightarrow h = \sqrt {2500 - {x^2}} \\\\S(x) = \frac{1}{2} \times 2x \times h = x\sqrt {2500 - {x^2}} \;\;\;D = \left[ {0\;,\;50} \right]\\\\S'(x) = \sqrt {2500 - {x^2}} + x \times \frac{{ - 2x}}{{2\sqrt {2500 - {x^2}} }}\\\\ = \frac{{2500 - {x^2} - {x^2}}}{{\sqrt {2500 - {x^2}} }} = \frac{{2500 - 2{x^2}}}{{\sqrt {2500 - {x^2}} }}\\\\\mathop \Rightarrow \limits^{S'(x) = 0} \;\;2500 - 2{x^2} = 0 \Rightarrow {x^2} = \frac{{2500}}{2} = 1250\\\\ \Rightarrow x = 25\sqrt 2 \\\\\mathop \Rightarrow \limits^{h = \sqrt {2500 - {x^2}} } \;\;h = \sqrt {2500 - 1250} = \sqrt {1250} \\\\ \Rightarrow h = 25\sqrt 2 \\\\S = x \times h = 25\sqrt 2 \times 25\sqrt 2 \Rightarrow S = 1250\end{array}\)

ب با توجه \(S = \frac{1}{2} \times 50 \times 50 \times \sin \theta \) بیشترین مساحت وقتی است که \(\sin \theta = 1\) باشد، پس \(\theta = {90^ \circ }\) می شود، پس خواهیم داشت:

\(S = \frac{1}{2} \times 50 \times 50 \times \sin \theta = \frac{1}{2} \times 50 \times 50 \times 1 = 1250\)

\(\begin{array}{l}S(x) = 2xy = 2x(12 - {x^2}) = 24x - 2{x^3}\\\\S'(x) = 24 - 6{x^2}\;\;\mathop \Rightarrow \limits^{S'(x) = 0} \;\;24 - 6{x^2} = 0\\\\ \Rightarrow {x^2} = 4 \Rightarrow x = 2\\\\\mathop \Rightarrow \limits^{y = 12 - {x^2}} \;\;y = 12 - 4 = 8\end{array}\)

طول مستطیل برابر با 4 و عرض آن برابر با 8 است.

\(\begin{array}{l}S(x) = (x + 2)(y + 4)\;\;\mathop \Rightarrow \limits^{xy = 32} \;\;y = \frac{{32}}{x}\\\\ \Rightarrow S(x) = 4x + \frac{{64}}{x} + 40\\\\S'(x) = 4 - \frac{{64}}{{{x^2}}} = \frac{{4{x^2} - 64}}{{{x^2}}}\\\\\mathop \Rightarrow \limits^{S'(x) = 0} \;\;4{x^2} - 64 = 0 \Rightarrow {x^2} = 16 \Rightarrow \left\{ \begin{array}{l}x = 4\\\\y = 8\end{array} \right.\end{array}\)

ابعاد جعبه برابر است با \((4 + 2 = )\;6\) و \((8 + 4 = )\;12\)

\(\begin{array}{l}\left. \begin{array}{l}{t_1} = \frac{{200 - x}}{3}\\\\{t_2} = \frac{{\sqrt {3600 + {x^2}} }}{2}\end{array} \right\}\\\\ \Rightarrow t = {t_1} + {t_2} = \frac{{200 - x}}{3} + \frac{{\sqrt {3600 + {x^2}} }}{2}\\\\ = \frac{1}{6}(400 - 2x + 3\sqrt {3600 + {x^2}} )\\\\t' = \frac{1}{6}( - 2 + 3 \times \frac{{2x}}{{2\sqrt {3600 + {x^2}} }})\;\;\mathop \Rightarrow \limits^{t' = 0} \;\;2 = \frac{{3x}}{{\sqrt {3600 + {x^2}} }}\\\\ \Rightarrow 2\sqrt {3600 + {x^2}} = 3x\\\\\mathop \Rightarrow \limits^{{{(\;)}^2}} \;\;4(3600 + {x^2}) = 9{x^2} \Rightarrow 5{x^2} = 4 \times 3600\\\\ \Rightarrow {x^2} = 2880 \Rightarrow x = 24\sqrt 5 \\\\ \Rightarrow t = \frac{1}{6}(400 - 2 \times 24\sqrt 5 + 3\sqrt {3600 + 2880} ) = 100\end{array}\)