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SVETLANKA909090 [29]
3 years ago
9

Marianne has been collecting donations for her biscuit stall at the school summer fayre. There are some luxury gift tins of bisc

uits to be sold at £5 each, normal packets at £1 each, and mini-packs of 2 biscuits at 10p each. She tells Amy that she has recieved exactly 100 donations in total, with a collective value of £100, amd that her stock of £1 packets is very low compared with the other items. Amy wants to work out how many of each item Marianne has. Show how she can do it.
Mathematics
2 answers:
sergeinik [125]3 years ago
7 0
The problem conditions give rise to 2 equations in 3 unknowns. Let L, M, N represent the number of Luxury, Mini, and Normal packets sold.
.. L +M +N = 100 . . . . . . the number of packets sold
.. 5L +0.1M +N = 100 . . the value of donations
These result in the relationships
.. L = (9/40)M
.. N = 100 -(49/40)M

There are three integer solutions in which the numbers are non-negative.
.. (L, M, N) = (0, 0, 100) or (9, 40, 51) or (18, 80, 2)

If Marianne sold 100 normal packets, her stock would be "very low" compared to the others.

If Marianne sold 51 normal packets, her stock may be "very low" with respect to the others, depending on how many of each she started with. This might be the solution if we require non-zero numbers of all packets were sold.
Marta_Voda [28]3 years ago
6 0
I don't know what 10p is so before I answer I'd have to know what that is
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3 years ago
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Step-by-step explanation:


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2 years ago
A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea
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Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

The side made of heavy duty material of length of <em>x </em>costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are <em>x, y, y</em>.  Thus

C=4x+4y+4y+16x\\C=20x+8y

We know that the total area of rectangular field should be 2250 square yards,

x\cdot y=2250

We can say that y=\frac{2250}{x}

Substituting into the total cost of the rectangular field, we get

C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}

We have to figure out where the function is increasing and decreasing. Differentiating,

\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}

Next, we find the critical points of the derivative

20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30

Because the length is always positive the only point we take is x=30. We thus test the intervals (0, 30) and (30, \infty)

C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0

we see that total cost function is decreasing on (0, 30) and increasing on (30, \infty). Therefore, the minimum is attained at x=30, so the minimal cost is

C(30)=20(30)+\frac{18000}{30}\\C(30)=1200

The least cost of fencing for the rancher is $1200

Here’s the diagram:

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