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Bess [88]
3 years ago
7

Suppose you want to build a fish tank in the shape of a right rectangular box with square base and no top which will hold 6 cubi

c feet of water. the glass for the sides costs $1 per square foot, and the metal for the bottom costs $1.50 per square foot. what dimensions for the tank will minimize the cost?
Mathematics
1 answer:
AlekseyPX3 years ago
4 0
You must develop a cost function C(x) and then minimize its value.
How much dwill the glass cost?  It's $1 per sq ft, and the total area of the glass is 4(xh), where x is the length of one side of the base and h is the height of the tank.  The area of the metal bottom is x^2, which we must multiply by $1.50 per sq ft. 

This cost function will look like this:  C(x) = 4($1/ft^2)xh + ($1.50/ft^2)x^2

but we know that (x^2)h= 6 cu ft, or h = (6 cu ft) / (x^2).  Subst. this last result into the C(x) equation, immediately above:

C(x) = 4($1/ft^2)x[6 ft^3 / x^2] + ($1.50/ft^2)x^2

Let's focus on the numerical values and ditch the units of measurement for now:

C(x) = 4x(4/x^2) + 1.50x^2, or

C(x) = 16/x + 1.5x^2

Differentiate this with respect to x:

C '(x) = -16 / x^2 + 3 x

Set this equal to 0 and solve for x:    -16/x^2 = -3x, or 16 = 3x^3

Then x^3 = 16/3, and x = 5 1/3 ft.  We already have the formula  

(x^2)h= 6 cu ft, so if x = 5 1/3, or 16/3, then (16/3)^2 h = 6, or

h = 6 / [16/3]^2.

h = 6 (9/256) = 0.21 ft.  While possible, this h = 0.21 ft seems quite unlikely.


Please work through this problem yourself, making sure you understand each step.  If questions arise, or if you find an error in my approach, please let me know.

Once again:
1.  Write a formula for the total cost of the material used:  4 sides of dimensions xh each, plus 1 bottom, of dimensions x^2.  Include the unit prices:  $1 per square foot for the sides and $1.50 per square foot for the bottom.
2.  Differentiate C(x) with respect to x.
3.  Set C '(x) = 0 and solve for the critical value(s).
4.  Calculate h from your value for x.
5.  Write the dimensions of the tank:  bottom:  x^2; height:  h
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Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h
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Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

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3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

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