Transcript Document

8D Max and Min Problems When the Function is Unknown
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8D Max and Min Problems When the Function is Unknown
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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
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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
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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
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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
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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
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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
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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.
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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
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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
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