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# another math question watch

1. A student needs tuition for an exam. She may consult tutor A, who charges 50 euros an hour, ot tutor B, who charges 40 euros an hour. ( She is not required to take a whole number of hours of tuition from either tutor.)

She estimates that if she has x hours tuition from tutor A and y hours from tutor B, then she will be able to solve 4x+2y+2xy problems on the exam.( This needn't be a whole number) Her parents have kindly provided a sum of M euros for the tutoring.

A) What is the cost of her tution, in terms of x and y?

B) Assuming she spends all money on tuition, how many problems will she be able to solve, in terms of x and M?

C) In terms of M, what is the value of x enaling her to solve the most problems?
2. (Original post by Teatime)
A student needs tuition for an exam. She may consult tutor A, who charges 50 euros an hour, ot tutor B, who charges 40 euros an hour. ( She is not required to take a whole number of hours of tuition from either tutor.)

She estimates that if she has x hours tuition from tutor A and y hours from tutor B, then she will be able to solve 4x+2y+2xy problems on the exam.( This needn't be a whole number) Her parents have kindly provided a sum of M euros for the tutoring.

A) What is the cost of her tution, in terms of x and y?

B) Assuming she spends all money on tuition, how many problems will she be able to solve, in terms of x and M?

C) In terms of M, what is the value of x enaling her to solve the most problems?
This looks like linear programming or some sort of discrete maths...

A) X hours costs 50X euros, Y hours costs 40Y euros. So the cost would be C=50X+40Y I think...

B) Assuming all M is spent, she can solve M=4X+2Y+2XY problems.

I get stuck after this, sorry...
3. cost of tuition = 50x + 40y

parents give her M money, all of which spends on tuition

so M= 50x+40y

so y=(M-50x)/40

no. of problems solved = 4x + 2y + 2xy

replace y with (M-50x)/40 wherever it appears

no. of problems solved = 4x + 2(M-50x)/40 + 2x(M-50x)/40

= (80x+M-50x+MX-50x^2)/20

= (M+(30+M)x-50x^2)/20

this is no. of problems solved in terms of M and x

are you familiar with differentiation ?

We now need to differentiate with respect to x to find the turning point. We know this will be a maximum because the coefficient of the x^2 term is negative (this means the graph of the function will be an upside-down parabola, and you can see that the turning point of an upside-down parabola is its maximum point)

differential is (30+M-100x)/20

set (30+M-100x)/20 = 0

implies 30 + M - 100x = 0

implies x = (30+M)/100

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