Người ta cần làm một cái bồn chứa dạng hình trụ có thể tích 1000 lít bằng inox để chứa nước, tính bán kính R của hình trụ đó sao cho diện tích toàn phần của bồn chứa đạt giá trị nhỏ nhất:

Gọi h và R lần lượt là chiều cao và bán kính đáy (đơn vị: mét).

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9iaadIgacqaHapaCcaWGsbWaaWbaaSqabeaacaaIYaaaaOGaeyyp % a0JaaGymaiabgkziUkaadIgacqGH9aqpdaWcaaqaaiaaigdaaeaacq % aHapaCcaWGsbWaaWbaaSqabeaacaaIYaaaaaaaaaa!4631! V = h\pi {R^2} = 1 \to h = \frac{1}{{\pi {R^2}}}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWG0bGaamiCaaqabaGccqGH9aqpcaaIYaGaeqiWdaNaamOu % amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaikdacqaHapaCcaWGsb % GaamiAaiabg2da9iaaikdacqaHapaCcaWGsbWaaWbaaSqabeaacaaI % YaaaaOGaey4kaSIaaGOmaiabec8aWjaadkfadaWcaaqaaiaaigdaae % aacqaHapaCcaWGsbWaaWbaaSqabeaacaaIYaaaaaaakiabg2da9iaa % ikdacqaHapaCcaWGsbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSYaaS % aaaeaacaaIYaaabaGaamOuaaaadaqadaqaaiaadkfacqGH+aGpcaaI % WaaacaGLOaGaayzkaaaaaa!5D14! {S_{tp}} = 2\pi {R^2} + 2\pi Rh = 2\pi {R^2} + 2\pi R\frac{1}{{\pi {R^2}}} = 2\pi {R^2} + \frac{2}{R}\left( {R > 0} \right)\)

Cách 1: Khảo sát hàm số, thu được \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamOuaaGaayjkaiaawMcaamaaBaaaleaaciGGTbGaaiyAaiaa % c6gaaeqaaOGaeyi1HSTaamOuaiabg2da9maakeaabaWaaSaaaeaaca % aIXaaabaGaaGOmaiabec8aWbaaaSqaaiaaiodaaaGccqGHshI3caWG % ObGaeyypa0ZaaSaaaeaacaaIXaaabaGaeqiWda3aaOqaaeaadaWcaa % qaaiaaigdaaeaacaaI0aGaeqiWda3aaWbaaSqabeaacaaIYaaaaaaa % aeaacaaIZaaaaaaaaaa!5079! f{\left( R \right)_{\min }} \Leftrightarrow R = \sqrt[3]{{\frac{1}{{2\pi }}}} \Rightarrow h = \frac{1}{{\pi \sqrt[3]{{\frac{1}{{4{\pi ^2}}}}}}}\)

Cách 2: Dùng bất đẳng thức:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWG0bGaamiCaaqabaGccqGH9aqpcaaIYaGaeqiWdaNaamOu % amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaikdacqaHapaCcaWGsb % GaamiAaiabg2da9iaaikdacqaHapaCcaWGsbWaaWbaaSqabeaacaaI % YaaaaOGaey4kaSIaaGOmaiabec8aWjaadkfadaWcaaqaaiaaigdaae % aacqaHapaCcaWGsbWaaWbaaSqabeaacaaIYaaaaaaakiabg2da9iaa % ikdacqaHapaCcaWGsbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSYaaS % aaaeaacaaIXaaabaGaamOuaaaacqGHRaWkdaWcaaqaaiaaigdaaeaa % caWGsbaaaiabgwMiZkaaiodadaGcbaqaaiaaikdacqaHapaCcaWGsb % WaaWbaaSqabeaacaaIYaaaaOGaaiOlamaalaaabaGaaGymaaqaaiaa % dkfaaaGaaiOlamaalaaabaGaaGymaaqaaiaadkfaaaaaleaacaaIZa % aaaOGaeyypa0JaaG4mamaakeaabaGaaGOmaiabec8aWbWcbaGaaG4m % aaaaaaa!6CD9! {S_{tp}} = 2\pi {R^2} + 2\pi Rh = 2\pi {R^2} + 2\pi R\frac{1}{{\pi {R^2}}} = 2\pi {R^2} + \frac{1}{R} + \frac{1}{R} \ge 3\sqrt[3]{{2\pi {R^2}.\frac{1}{R}.\frac{1}{R}}} = 3\sqrt[3]{{2\pi }}\)

Dấu bằng xảy ra khi và chỉ khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuamaaCa % aaleqabaGaaG4maaaakiabg2da9maalaaabaGaaGymaaqaaiaaikda % cqaHapaCaaaaaa!3C08! {R^3} = \frac{1}{{2\pi }}\)