英文数学微积分题....高分求解!!!!!
1.Findthedimensionsofthelargestrectanglethatcanbeinscribedinthesemicircley=√4-x平方.2.a...
1. Find the dimensions of the largest rectangle that can be inscribed in the semicircle y=√4-x平方.
2. a closed cylindrical can is to have a capacity of 16pi cm3. what are the radius of the base and the height of the cylinder for the total surface area to be a minumum.
3. a company manufactures items at 2 dollars per item and sells then for x dollars per item. if the number sold is 800/x2 per month, find the value of x for which the company could expect to maximie its monthly profit.
4. A traveller employs a man to drive him from A to B for an hourly payment of P dollars. running costs of the car, which are also paid by the traveller, are Kv*3 dollars per hour, where vkm h*-1 is the speed, and k is a constant. find the uniform speed that will minimize the total cost of the journey.
5. if y=ax+b/x and if y=13 when x=1 and x=20 when x=2, find the values of a and b and the value of x for which y is a minimum.
感激不尽!!!!!!!!完成保证再加分!!!!
p.s 是求解,不是求翻译.谢谢. 展开
2. a closed cylindrical can is to have a capacity of 16pi cm3. what are the radius of the base and the height of the cylinder for the total surface area to be a minumum.
3. a company manufactures items at 2 dollars per item and sells then for x dollars per item. if the number sold is 800/x2 per month, find the value of x for which the company could expect to maximie its monthly profit.
4. A traveller employs a man to drive him from A to B for an hourly payment of P dollars. running costs of the car, which are also paid by the traveller, are Kv*3 dollars per hour, where vkm h*-1 is the speed, and k is a constant. find the uniform speed that will minimize the total cost of the journey.
5. if y=ax+b/x and if y=13 when x=1 and x=20 when x=2, find the values of a and b and the value of x for which y is a minimum.
感激不尽!!!!!!!!完成保证再加分!!!!
p.s 是求解,不是求翻译.谢谢. 展开
3个回答
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1.内接圆,那个y与x的关系式看不明白。
2.设定底面半径为r, 圆柱高为h
v=16pi=pi*r^2*h
表面积s=2*pi*r^2+(pi*2*r)*h有最小值,
s=2*pi*(r^2+r*h),将h用r替换,再取导数,求极值,r=2。
3.每个item的盈利为(x-2)
销售量为800/(x^2),我认为应该是x的平方
盈利m=每个item的盈利*销售量=(x-2)*(800/(x^2)), 再求导数,取极值就行
4. 设定A到B的距离为D km
man的报酬是P d/h
车速是 v km/h
车的消耗仍由traveller支付,消耗为 Kv*3 d/h (我有点怀疑这个是不是v的3次方)
A到B的时间 t = D/v (h)
消耗 m = t*(P + Kv*3)
再求导数,取极值就行
5. 求得a,b的值,使得y有最小值
y= ax+b/x >= 2*t , t是根号下a*b的值
只要取得a*b的最小值即可 (这里有点问题吧, y=13 when x=1 and x=20 when x=2,应该是y=20吧)
y=13=a+b
y=20=2*a+b/2
联立求解,可得a,b的值
2.设定底面半径为r, 圆柱高为h
v=16pi=pi*r^2*h
表面积s=2*pi*r^2+(pi*2*r)*h有最小值,
s=2*pi*(r^2+r*h),将h用r替换,再取导数,求极值,r=2。
3.每个item的盈利为(x-2)
销售量为800/(x^2),我认为应该是x的平方
盈利m=每个item的盈利*销售量=(x-2)*(800/(x^2)), 再求导数,取极值就行
4. 设定A到B的距离为D km
man的报酬是P d/h
车速是 v km/h
车的消耗仍由traveller支付,消耗为 Kv*3 d/h (我有点怀疑这个是不是v的3次方)
A到B的时间 t = D/v (h)
消耗 m = t*(P + Kv*3)
再求导数,取极值就行
5. 求得a,b的值,使得y有最小值
y= ax+b/x >= 2*t , t是根号下a*b的值
只要取得a*b的最小值即可 (这里有点问题吧, y=13 when x=1 and x=20 when x=2,应该是y=20吧)
y=13=a+b
y=20=2*a+b/2
联立求解,可得a,b的值
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1. It's clear that the 4 vertices are A(a, 0), B(a, √(4 - a²)), C(-a, √(4 - a²)), D(-a, 0)
Its area S = AD*AB = 2a*√(4 - a²)
S' = 2√(4 - a²) + 2a*(1/2)(-2a)/√(4 - a²) = 2√(4 - a²) - 2a²/√(4 - a²) = 0
√(4 - a²) = a²/√(4 - a²)
4 - a² = a²
a = √2
dimension: length = 2a = 2√2; width = √(4 - a²) = √2
2. Assume the radius of the base is r cm, height is h cm. V = πr²h = 16π, h = 16/r²
The total surface S = 2πr² + 2πrh = 2π(r² + rh) = 2π(r² + r*16/r²) = 2π(r² + 16/r)
S' = 2π(2r - 16/r²) = 0
r³ = 8
r = 2
3. The profit from each item is x - 2 dollars, and the total profit in a month isP = (x-2)*800/x² dollars
P' = 800(1/x² + (x-2)(-2)/x³)
= 800(x - 2x + 4)/x³ = 0
x = 4
4. Assume the distance between A and B is d km, a constant. Then the time needed to reach B from A is d/v hours, the total expense is:
E = Pd/v + (Kv³)(d/v)
= d(P/v + Kv²)
E' = d(-P/v² + 2Kv) = 0
v³ = P/(2K)
V = ³√(P/(2K)) (cubic root)
5. There's a minor error, should be "y = 20 when x = 2".
x = 1, y = 13: a + b = 13 (1)
x = 2, y = 20: 2a + b/2 = 20 (2)
From (1) and (2): b = 4, a = 9
y = 9x + 4/x
y' = 9 - 4/x² = 0
x = ±2/3
Its area S = AD*AB = 2a*√(4 - a²)
S' = 2√(4 - a²) + 2a*(1/2)(-2a)/√(4 - a²) = 2√(4 - a²) - 2a²/√(4 - a²) = 0
√(4 - a²) = a²/√(4 - a²)
4 - a² = a²
a = √2
dimension: length = 2a = 2√2; width = √(4 - a²) = √2
2. Assume the radius of the base is r cm, height is h cm. V = πr²h = 16π, h = 16/r²
The total surface S = 2πr² + 2πrh = 2π(r² + rh) = 2π(r² + r*16/r²) = 2π(r² + 16/r)
S' = 2π(2r - 16/r²) = 0
r³ = 8
r = 2
3. The profit from each item is x - 2 dollars, and the total profit in a month isP = (x-2)*800/x² dollars
P' = 800(1/x² + (x-2)(-2)/x³)
= 800(x - 2x + 4)/x³ = 0
x = 4
4. Assume the distance between A and B is d km, a constant. Then the time needed to reach B from A is d/v hours, the total expense is:
E = Pd/v + (Kv³)(d/v)
= d(P/v + Kv²)
E' = d(-P/v² + 2Kv) = 0
v³ = P/(2K)
V = ³√(P/(2K)) (cubic root)
5. There's a minor error, should be "y = 20 when x = 2".
x = 1, y = 13: a + b = 13 (1)
x = 2, y = 20: 2a + b/2 = 20 (2)
From (1) and (2): b = 4, a = 9
y = 9x + 4/x
y' = 9 - 4/x² = 0
x = ±2/3
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大神!!!!还差最后一道题....完了再加50分!!!!拜托拜托了!!!!
a printed page of a book is to have side margins of 1cm, a top margin of 2cm and a bottom margin of 3cm. it is to contain 200cm2 of printed matter. find the dimensions of the page if the area of paper used is to be a minimum.
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追答
Assume the length of the printed area is x cm, its width is 200/x cm.
The length of the whole page is x + 2 + 3 = x + 5 cm; the width 200/x + 1 + 1 = 2 + 200/x cm
The area of the pater is S = (x+5)(2 + 200/x)
= 2x + 10 + 200 + 1000/x
= 2x + 1000/x + 210
S' = 2 -1000/x² = 0
x² = 500
x = 10√5 cm
The length of the whole page is x + 5 = 10√5 + 5 cm; width = 2+200/x = 2 + 4√5 cm
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1.的翻译是求半圆y=√4-x平方的内接矩形最大面积
答案是4
答案是4
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