SOLUTION: A machine comes in 2 parts, which weigh 'x' kg and 'b' kg respectively. The cost 'c' of the machine is given by c = 2x + b. The earning capacity 'y' of the machine is given by y =

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A machine comes in 2 parts, which weigh 'x' kg and 'b' kg respectively. The cost 'c' of the machine is given by c = 2x + b. The earning capacity 'y' of the machine is given by y =       Log On


   



Question 1012464: A machine comes in 2 parts, which weigh 'x' kg and 'b' kg respectively. The cost 'c' of the machine is given by c = 2x + b. The earning capacity 'y' of the machine is given by y = x(x+b). If 'c' has the fixed value 10, express y as a function of x and hence find the value of x for which y is a maximum. Find the maximum value of y.
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A machine comes in 2 parts, which weigh 'x' kg and 'b' kg respectively. The cost 'c' of the machine is given by c = 2x + b. The earning capacity 'y' of the machine is given by y = x(x+b). If 'c' has the fixed value 10, express y as a function of x and hence find the value of x for which y is a maximum. Find the maximum value of y.

c = cost = 2x + b.

since c = cost = 10, you get 10 = 2x + b.

solve for b to get b = 10 - 2x.

y = x * (x+b)

since b = 10 - 2x, you get y = x * (x + 10 - 2x)

simplify to get y = x * (10 - x)

simplify further to get y = 10x - x^2

set this equation into standard quadratic form of 0 = ax^2 + bx + o get y = -x^2 + 10x.

this means that:
a = -1
b = 10
c = 0

the x value of the max/min point of a quadratic equation is at x = -b/2a.

this becomes x = -10/-2 which results in x = 5.

when x = 5, y = -x^2 + 10x becomes y = -25 + 50 which becomes y = 25.

the maximum earning capacity is therefore 25 units of whatever denomination you are using.

you can graph this equation and it will show you the same result visually.

look below:

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