A dressmaker has purchased a 150ml roll of material to make

A dressmaker has purchased a 150ml roll of material to make some tops. The fabric is 1mm thick. The roll has an inner radius of 10 cm.

To answer this question you must construct a mathematical model, defining all variables you introduce and state any assumptions vou make A dressmaker has purchased a 150m roll of material to make some tops. The fabric is 1mm thick. The roll has an inner radius of 10cm a) Calculate the outer radius of the roll b) When the fabric was delivered the roll was sat in some oil, which has soaked into the first 2cm of the outside of the roll, ruining the fabric. Estimate the length of undamaged fabric (c) The dressmaker is preparing for a market day. They know from previous experience that if they charge $50 per top they will sell 40 tops. If they charge $70 per top they will onlv sell 20 The material costs $8 per metre. The dressmaker needs 2.5 metres for each top. The rest of the materials cost $2 per top. Assume the dressmaker only has to pay for the materials used in the tops sold (i) Find the quadratic function which models the dressmaker\'s profit (ii) Maximise the function from 4(c)i to determine what price the dressmaker should charge per top to maximise their profit

Solution

Outer radius will be (10+n*(0.1)) where n = number of windings.

So sum of all the lengths = perimeter which shoulb be equal to length of the roll. Since all are in A.p.

So 15000 = 2pi*n/2(10+ 10+n*(0.1))

n = 140.305 So it took just more than 140 windings.

So outer radius will be 0.1(141) +10 = 24.1 cm

As we have seen that length of roll depends on L = 2*pi*n/2 (10 + 10+0.1(n))

where n = number of windings and 10+(0.1)n = outer radius.

Since last 2 cm is damaged, 24.1-2 = 22.1cm

10 + (0.1)n = 22.1

n = 122

So roll length till 122 windings = pi*122(22.1+10) = 123.03m

(c) cost per metre = 8

Cost per top = 2.5*8+2 = 22$

Profit = number of tops * profit/top

Profit per top is related to quantity sold linearly which relates to price as x + price = 90

So price = 90 -x

= x* (90-x-22).

= 68x - x^2

Price to maximize is when x = 34