若△abc三边长a.b.c均为整数,且1\a+1\b+3\ab=4\1,a+b-c=8,设△abc的面积为S,则S最大值是?最小值是?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/04 19:47:42
若△abc三边长a.b.c均为整数,且1\a+1\b+3\ab=4\1,a+b-c=8,设△abc的面积为S,则S最大值是?最小值是?

若△abc三边长a.b.c均为整数,且1\a+1\b+3\ab=4\1,a+b-c=8,设△abc的面积为S,则S最大值是?最小值是?
若△abc三边长a.b.c均为整数,且1\a+1\b+3\ab=4\1,a+b-c=8,设△abc的面积为S,则S最大值是?最小值是?

若△abc三边长a.b.c均为整数,且1\a+1\b+3\ab=4\1,a+b-c=8,设△abc的面积为S,则S最大值是?最小值是?
根据题意有:
1/a+1/b+3/ab=1/4
即:(a+b+3)/ab=1/4
所以:
ab=4(a+b+3).(1)
又a+b-c=8
所以c=(a+b)-8.(2)
cosc=(a^2+b^2-c^2)/(2ab)
={a^2+b^2-[(a+b)-8]^2}/[8(a+b-3)]
=(a+b-11)/(a+b+3) 令a+b=t,则有:
cosc=(t-11)/(t+3)
所以三角形的面积s=(1/2)absinc=(1/2)*4*(t+3)√[(t+3)^2-(t-11)^2]/(t+3)=4√[7(t-4)].
把(1)变形得到:
(a – 4)(b – 4) = 28 = 1*28 =4*7
所以tmax=5+32=37,tmin=8+11=19.
所以smax=4 √[7*(37-4)]=4√231;
sminx=4√[7*(19-4)]=4√105.

Smax=4√231,Smin=4√105.

1/a + 1/b + 3/ab = 1/4 => 4b + 4a + 12 = ab => ab – 4a – 4b = 12 => (a – 4)(b – 4) = 28 = 1*28 = 2*14 = 4*7 = 7*4 = 14*2 = 28*1 ,分别可以解得 (a,b,c) = (5,32,29),(6,18,16),(8,11,11),(11,8,11),(18,6,16),(32,...

全部展开

1/a + 1/b + 3/ab = 1/4 => 4b + 4a + 12 = ab => ab – 4a – 4b = 12 => (a – 4)(b – 4) = 28 = 1*28 = 2*14 = 4*7 = 7*4 = 14*2 = 28*1 ,分别可以解得 (a,b,c) = (5,32,29),(6,18,16),(8,11,11),(11,8,11),(18,6,16),(32,5,29)。
一共有三个面积,由余弦定理和正弦定理可得,
cosA = (52 + 322 - 292)/(2 * 5 * 32) = 13/20 => S1 = (1/2) * 5 * 32sinA = 80 √231/20 = 4 √231 。
cosA = (62 + 182 – 162)/(2 * 6 * 18) = 13/27 => S2 = (1/2) * 6 * 18sinA = 54 * 4√35/27 = 8√35 。
cosA = (82 + 112 - 112)/(2 * 8 * 11) = 4/11 => S3 = (1/2) * 8 * 11sinA = 44√105/11 = 4√105 。
综上所述,S的最大值是4 √231 ,最小值是4√105 。
选我吧!!

收起

1/a + 1/b + 3/ab = 1/4 => 4b + 4a + 12 = ab => ab – 4a – 4b = 12 => (a – 4)(b – 4) = 28 = 1*28 = 2*14 = 4*7 = 7*4 = 14*2 = 28*1 ,分别可以解得 (a,b,c) = (5,32,29),(6,18,16),(8,11,11),(11,8,11),(18,6,16),(32,...

全部展开

1/a + 1/b + 3/ab = 1/4 => 4b + 4a + 12 = ab => ab – 4a – 4b = 12 => (a – 4)(b – 4) = 28 = 1*28 = 2*14 = 4*7 = 7*4 = 14*2 = 28*1 ,分别可以解得 (a,b,c) = (5,32,29),(6,18,16),(8,11,11),(11,8,11),(18,6,16),(32,5,29)。
一共有三个面积,由余弦定理和正弦定理可得,
cosA = (52 + 322 - 292)/(2 * 5 * 32) = 13/20 => S1 = (1/2) * 5 * 32sinA = 80 √231/20 = 4 √231 。
cosA = (62 + 182 – 162)/(2 * 6 * 18) = 13/27 => S2 = (1/2) * 6 * 18sinA = 54 * 4√35/27 = 8√35 。
cosA = (82 + 112 - 112)/(2 * 8 * 11) = 4/11 => S3 = (1/2) * 8 * 11sinA = 44√105/11 = 4√105 。
综上所述,S的最大值是4 √231 ,最小值是4√105 。

收起

1\a+1\b+3\ab=4\1 ?是1/4吧,求原题!!!

54. 180;96