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saul85 [17]
3 years ago
6

Show that the sum of two concave functions is concave. Is the product of two concave functions also concave?

Mathematics
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

1.

Let f and g concave functions. Then for a in (0,1), f( (1-\lambda)x + \lambda y ) \geq (1-\lambda)f(x) + \lambda f(y) and g( (1-\lambda)x + \lambda y ) \geq (1-\lambda)g(x) + \lambda g(y)

Now, (f+g)( (1-\lambda)x + \lambda y ) = f( (1-\lambda)x + \lambda y ) + g( (1-\lambda)x + \lambda y )\geq (1-\lambda)f(x) + \lambda f(y) + (1-\lambda)g(x) + \lambda g(y) = (1-\lambda)(f(x)+g(x)) + \lambda (f(x)+g(x))= (1-\lambda)(f+g)(x) + \lambda(f+g)(x)

This shows that the function f+g is concave.

2. Let f and g concave functions.

(fg)( (1-a)x + \lambda y )= f( (1-\lambda)x + \lambda y )g( (1-\lambda)x + \lambda y )  \geq [(1-a)f(x) + \lambda f(y)] [(1-\lambda)g(x) + \lambda g(y)] = (1-\lambda)^2 f(x)g(x) + \lambda (1-\lambda) f(x)g(y)+ \lambda (1-\lambda) f(y)g(x)+ \lambda ^2 f(y)g(y) \neq (1-\lambda)f(x)g(x) + \lambda f(y)g(y)

This shows that the product of two concave functions isn't concave.

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2. Sue is buying a 13-pound mixture of gummy candy, jelly beans, and hard candy. The cost of gummy candy is $1.20 per pound, jel
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Answer:

Store used 7.5 pounds of gummy candy, 2.5 pounds of jelly beans, and 3 pounds of hard candy.

Step-by-step explanation:

Let the amount of gummy candy be 'x'.

Let the amount of jelly beans be 'y'.

Let the amount of hard candy 'z'.

Now Given:

Sue is buying 13 pound of mixture.

So we can say that;

x+y+z =13

But Given:

The mixture calls for three times as many gummy candy pieces as jelly beans.

x=3y

Substituting the value of x in above equation we get;

3y+y+z=13\\\\4y +z =13 \ \ \ \ \ equation \ 1

Also Given:

cost of gummy candy = $1.20

cost of jelly beans = $2.00

cost of hard candy = $2.60

Total Cost of mixture  = $21.80

Now Total Cost of mixture is equal to cost of gummy candy multiplied amount of gummy candy plus cost of jelly bean multiplied amount of jelly bean plus cost of hard candy multiplied amount of hard candy.

framing in equation form we get;

1.2x+2y+2.6z=21.80

But  x=3y

So 1.2(3y)+2y+2.6z=21.80\\\\3.6y+2y+2.6z=21.80\\\\5.6y+2.6z=21.80

Now Multiplying by both side by 10 we get;

10(5.6y+2.6z)=21.80\times 10\\\\10\times5.6y + 10\times2.6z =218\\\\56y+26z=218 \ \ \ \ \ equation \ 2

Now Multiplying equation 1 by 14 we get;

14(4y +z) =13\times14\\\\14\times4y +14z =182\\\\56y+14z=182

Now Subtracting equation 3 from equation 2 we get;

(56y+26z) - (56y+14z) =218-182\\\\56y +26z-56y-14z=36\\\\12z=36\\\\z=\frac{36}{12} = 3 \ pounds

Now Substituting value of z in equation 1 we get;

4y+z=13\\\\4y+3=13\\\\4y=13-3\\\\4y = 10\\\\y=\frac{10}{4} = 2.5 \ pounds

Now also;

x= 3y\\\\x =3\times2.5 =7.5 \ pounds

Hence Store used 7.5 pounds of gummy candy, 2.5 pounds of jelly beans, and 3 pounds of hard candy.

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