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 ar

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 ar
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.
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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 ar
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