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8D Max and Min Problems When the Function is Unknown 1 8D Max and Min Problems When the Function is Unknown 2 8D Max and Min Problems When the Function is Unknown 𝐓𝐡𝐞 𝐬𝐮𝐦 𝐨𝐟 𝟐 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐢𝐬 𝟏𝟎. 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐢𝐟 𝐭𝐡𝐞 𝐬𝐮𝐦 𝐨𝐟 𝐭𝐡𝐞𝐢𝐫 𝐬𝐪𝐮𝐚𝐫𝐞𝐬 𝐢𝐬 𝐚 𝐦𝐢𝐧𝐢𝐦𝐮𝐦 𝐥𝐞𝐭 𝒙 𝐫𝐞𝐩𝐫𝐞𝐬𝐞𝐧𝐭 𝐨𝐧𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐚𝐧𝐝 𝒚 𝐫𝐞𝐩𝐫𝐞𝐬𝐞𝐧𝐭 𝐭𝐡𝐞 𝐨𝐭𝐡𝐞𝐫 𝒙 + 𝒚 = 𝟏𝟎 𝒚 = 𝟏𝟎 − 𝒙 𝑺 𝒙 = 𝒙 𝟐 + 𝒚𝟐 = 𝒙𝟐 + (𝟏𝟎 − 𝒙)𝟐 = 𝒙𝟐 + 𝟏𝟎𝟎 − 𝟐𝟎𝒙 + 𝒙𝟐 𝑺 𝒙 = 𝟐𝒙𝟐 − 𝟐𝟎𝒙 + 𝟏𝟎𝟎 𝑺′ 𝒙 = 𝟒𝒙 − 𝟐𝟎 𝟎 = 𝟒𝒙 − 𝟐𝟎 𝒙=𝟓 𝟓 + 𝒚 = 𝟏𝟎 𝒚=𝟓 𝐓𝐡𝐞 𝟐 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐰𝐡𝐢𝐜𝐡 𝐠𝐢𝐯𝐞 𝐚 𝐦𝐢𝐧𝐦𝐮𝐦 𝐨𝐟 𝐭𝐡𝐞𝐢𝐫 𝐬𝐪𝐮𝐚𝐫𝐞𝐬 𝐚𝐫𝐞 𝐛𝐨𝐭𝐡 𝟓 3 8D Max and Min Problems When the Function is Unknown 𝐀 𝐜𝐮𝐛𝐨𝐢𝐝 𝐜𝐨𝐧𝐭𝐚𝐢𝐧𝐞𝐫 𝐰𝐢𝐭𝐡 𝐚 𝐛𝐚𝐬𝐞 𝐥𝐞𝐧𝐠𝐭𝐡 𝐭𝐰𝐢𝐜𝐞 𝐢𝐭𝐬 𝐰𝐢𝐝𝐭𝐡 𝐢𝐬 𝐦𝐚𝐝𝐞 𝐟𝐫𝐨𝐦 𝟒𝟖𝐦𝟐 𝐨𝐟 𝐦𝐞𝐭𝐚𝐥. 𝐚. 𝐬𝐡𝐨𝐰 𝐭𝐡𝐚𝐭 𝐭𝐡𝐞 𝐡𝐞𝐢𝐠𝐡𝐭 𝐢𝐬 𝐠𝐢𝐯𝐞𝐧 𝐛𝐲 𝟖 𝟐𝒙 𝒉= − 𝒙 𝟑 𝐓𝐒𝐀 = 𝟒𝟖𝐦𝟐 𝐓𝐒𝐀 = 𝟐 𝒉 × 𝒙 + 𝒙 × 𝟐𝒙 + 𝟐𝒙 × 𝒉 = 𝟐 𝒙𝒉 + 𝟐𝒙𝟐 + 𝟐𝒙𝒉 = 𝟒𝒙𝟐 + 𝟔𝒙𝒉 𝟒𝟖 = 𝟒𝒙𝟐 + 𝟔𝒙𝒉 𝟒𝟖 = 𝟒𝒙𝟐 + 𝟔𝒙𝒉 𝟔𝒙𝒉 = 𝟒𝟖 − 𝟒𝒙𝟐 𝟒𝟖 𝟒𝒙𝟐 𝒉= − 𝟔𝒙 𝟔𝒙 𝒉= 𝟖 𝟐𝒙 − 𝐚𝐬 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝 𝒙 𝟑 4 8D Max and Min Problems When the Function is Unknown 𝐀 𝐜𝐮𝐛𝐨𝐢𝐝 𝐜𝐨𝐧𝐭𝐚𝐢𝐧𝐞𝐫 𝐰𝐢𝐭𝐡 𝐚 𝐛𝐚𝐬𝐞 𝐥𝐞𝐧𝐠𝐭𝐡 𝐭𝐰𝐢𝐜𝐞 𝐢𝐭𝐬 𝐰𝐢𝐝𝐭𝐡 𝐢𝐬 𝐦𝐚𝐝𝐞 𝐟𝐫𝐨𝐦 𝟒𝟖𝐦𝟐 𝐨𝐟 𝐦𝐞𝐭𝐚𝐥. 𝐚. 𝐬𝐡𝐨𝐰 𝐭𝐡𝐚𝐭 𝐭𝐡𝐞 𝐡𝐞𝐢𝐠𝐡𝐭 𝐢𝐬 𝐠𝐢𝐯𝐞𝐧 𝐛𝐲 𝟖 𝟐𝒙 𝒉= − 𝒙 𝟑 𝐛. 𝐞𝐱𝐩𝐫𝐞𝐬𝐬 𝐭𝐡𝐞 𝐯𝐨𝐥𝐮𝐦𝐞, 𝑽, 𝐢𝐧 𝐭𝐞𝐫𝐦𝐬 𝐨𝐟 𝒙. 𝑽 = 𝒙 × 𝟐𝒙 × 𝒉 𝑽 = 𝟐𝒙 𝟐 𝟖 𝟐𝒙 − 𝒙 𝟑 𝟒𝒙𝟑 𝑽 = 𝟏𝟔𝒙 − 𝟑 5 8D Max and Min Problems When the Function is Unknown 𝐀 𝐜𝐮𝐛𝐨𝐢𝐝 𝐜𝐨𝐧𝐭𝐚𝐢𝐧𝐞𝐫 𝐰𝐢𝐭𝐡 𝐚 𝐛𝐚𝐬𝐞 𝐥𝐞𝐧𝐠𝐭𝐡 𝐭𝐰𝐢𝐜𝐞 𝐢𝐭𝐬 𝐰𝐢𝐝𝐭𝐡 𝐢𝐬 𝐦𝐚𝐝𝐞 𝐟𝐫𝐨𝐦 𝟒𝟖𝐦𝟐 𝐨𝐟 𝐦𝐞𝐭𝐚𝐥. 𝐚. 𝐬𝐡𝐨𝐰 𝐭𝐡𝐚𝐭 𝐭𝐡𝐞 𝐡𝐞𝐢𝐠𝐡𝐭 𝐢𝐬 𝐠𝐢𝐯𝐞𝐧 𝐛𝐲 𝟖 𝟐𝒙 𝒉= − 𝒙 𝟑 𝐛. 𝐞𝐱𝐩𝐫𝐞𝐬𝐬 𝐭𝐡𝐞𝐯𝐨𝐥𝐮𝐦𝐞, 𝑽, 𝐢𝐧 𝐭𝐞𝐫𝐦𝐬 𝐨𝐟 𝒙. 𝐜. 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐯𝐨𝐥𝐮𝐦𝐞 𝟒𝒙𝟑 𝑽(𝒙) = 𝟏𝟔𝒙 − 𝟑 𝟏𝟐𝒙𝟐 ′ 𝑽 (𝒙) = 𝟏𝟔 − 𝟑 = 𝟏𝟔 − 𝟒𝒙𝟐 𝑽′ 𝒙 = 𝟎 𝐟𝐨𝐫 𝐦𝐚𝐱 𝐨𝐫 𝐦𝐢𝐧 𝟎 = 𝟏𝟔 − 𝟒𝒙𝟐 𝟒 = 𝒙𝟐 𝒙 = ±𝟐 + 𝟎 − 𝐓𝐡𝐞 𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐯𝐨𝐥𝐮𝐦𝐞 𝐢𝐬 𝐚𝐜𝐡𝐢𝐞𝐯𝐞𝐝 𝐰𝐡𝐞𝐧 𝒙 = 𝟐 𝟒 × 𝟐𝟑 𝑽 𝒙 = 𝟏𝟔 × 𝟐 − 𝟑 𝟑𝟐 = 𝟑𝟐 − 𝟑 𝟏 = 𝟐𝟏 𝟑 𝐓𝐡𝐞 𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐯𝐨𝐥𝐮𝐦𝐞 𝐢𝐬 𝟐𝟏 𝟏 𝟑 𝐦 𝟑 6 8D Max and Min Problems When the Function is Unknown Find the minimum distance from the straight line with equation y = x – 4 to the point (1,1). 𝒅 𝒙 = 𝒙−𝟏 𝟐 + 𝒙−𝟒−𝟏 𝐥𝐞𝐭 𝒙𝟏 = 𝟏 𝐚𝐧𝐝 𝒚𝟏 = 𝟏 = 𝒙−𝟏 𝟐 + 𝒙−𝟓 𝐚𝐧𝐝 𝒙𝟐 = 𝒙 𝐚𝐧𝐝 𝒚𝟐 = 𝒚 = 𝒅 𝒙 = 𝒙𝟐 − 𝒙𝟏 𝟐 + 𝒚𝟐 − 𝒚𝟏 𝐎𝐑 𝒙𝟐 = 𝒙 𝐚𝐧𝐝 𝒚𝟐 = 𝒙 − 𝟒 𝟐 𝒅 𝒙 = 𝟐 𝟐 𝒙𝟐 − 𝟐𝒙 + 𝟏 + 𝒙𝟐 − 𝟏𝟎𝒙 + 𝟐𝟓 𝟐𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟔 7 8D Max and Min Problems When the Function is Unknown Find the minimum distance from the straight line with equation y = x – 4 to the point (1,1). 𝒅 𝒙 = = 𝐈𝐟 𝒇 𝒙 = 𝒈(𝒙) 𝟐𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟔 𝟐𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟔 𝟏 𝟐 𝒅′ 𝒏 𝐭𝐡𝐞𝐧 𝒇′ 𝒙 = 𝒏 𝒈(𝒙) 𝟏 𝒙 = 𝟐𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟔 𝟐 −𝟏 𝟐 𝒏−𝟏 × 𝒈′ (𝒙) × 𝟒𝒙 − 𝟏𝟐 𝟒𝒙 − 𝟏𝟐 = 𝟐 𝟐𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟔 𝟏 𝟐 8 8D Max and Min Problems When the Function is Unknown Find the minimum distance from the straight line with equation y = x – 4 to the point (1,1). 𝒅 𝒙 = 𝟐𝒙𝟐 𝒅′ (𝒙) − 𝟏𝟐𝒙 + 𝟐𝟔 𝟒𝒙 − 𝟏𝟐 = 𝟐 − 𝒅′ 𝒙 = 𝟎 𝐟𝐨𝐫 𝐦𝐚𝐱 𝐨𝐫 𝐦𝐢𝐧 𝟒𝒙 − 𝟏𝟐 𝟐 𝟐𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟔 𝟏 𝟐 =𝟎 𝟒𝒙 − 𝟏𝟐 = 𝟎 𝒙=𝟑 𝟐𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟔 𝟏 𝟐 + 𝐬𝐨 𝒙 = 𝟑 𝐢𝐬 𝐭𝐡𝐞 𝐦𝐢𝐧𝐢𝐦𝐮𝐦 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞. 𝒅 𝟑 = 𝟐(𝟑)𝟐 −(𝟏𝟐 × 𝟑) + 𝟐𝟔 𝒅= 𝟖 =𝟐 𝟐 ≈ 𝟐. 𝟖𝟑 𝐓𝐡𝐞 𝐦𝐢𝐧𝐢𝐦𝐮𝐦 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐬 𝐚𝐩𝐩𝐫𝐨𝐱𝐢𝐦𝐚𝐭𝐞𝐥𝐲 𝟐. 𝟖𝟑 𝐮𝐧𝐢𝐭𝐬. 9 8D Max and Min Problems When the Function is Unknown 1238 A cylinder of cheese is to be removed from a spherical piece of cheese of radius 8cm. What is the maximum volume of the cylinder of cheese? (answer to nearest whole unit) 𝟐𝒓 𝟏𝟔𝒄𝒎 𝟏𝟔𝟐 = (𝟐𝒓)𝟐 +𝒉𝟐 𝟒𝒓𝟐 = 𝟏𝟔𝟐 − 𝒉𝟐 𝒓𝟐 𝟐𝟓𝟔 − 𝒉𝟐 = 𝟒 𝟐 𝒉 𝒓𝟐 = 𝟔𝟒 − 𝟒 𝒉 𝐕𝐨𝐥𝐮𝐦𝐞 = 𝝅𝒓𝟐 𝐡 𝒉𝟐 = 𝝅 𝟔𝟒 − 𝐡 𝟒 𝒉𝟑 𝝅 𝐕𝐨𝐥𝐮𝐦𝐞 = 𝟔𝟒𝝅𝒉 − 𝟒 10 8D Max and Min Problems When the Function is Unknown 1238 A cylinder of cheese is to be removed from a spherical piece of cheese of radius 8cm. What is the maximum volume of the cylinder of cheese? (answer to nearest whole unit) 𝟐𝒓 𝟐𝟓𝟔 − 𝒉𝟐 𝒓 = 𝟒 𝟐 𝟏𝟔𝒄𝒎 𝒉 𝟐𝟓𝟔 − 𝟗. 𝟐𝟒𝟐 = 𝟒 𝒓𝟐 = 𝟒𝟐. 𝟔 𝒓𝟐 𝒉𝟑 𝝅 𝑽 = 𝟔𝟒𝝅𝒉 − 𝟒 𝒅𝑽 𝟑𝒉𝟐 𝝅 = 𝟔𝟒𝝅 − 𝒅𝒉 𝟒 𝒅𝑽 = 𝟎 𝐟𝐨𝐫 𝐦𝐚𝐱 𝐨𝐫 𝐦𝐢𝐧 𝒅𝒉 𝟑𝒉𝟐 𝝅 𝟎 = 𝟔𝟒𝝅 − 𝟒 𝟐 𝟑𝒉 𝟔𝟒 = 𝟒 𝟔𝟒 × 𝟒 𝒉𝟐 = 𝟑 𝒉= 𝐕𝐨𝐥𝐮𝐦𝐞 = 𝝅𝒓𝟐 𝐡 = 𝝅 × 𝟒𝟐. 𝟔 × 𝟗. 𝟐𝟒 ≈ 𝟏𝟐𝟑𝟖𝐜𝐦𝟑 𝟐𝟓𝟔 ≈ 𝟗. 𝟐𝟒 𝟑 11