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# Applied Maximum and Minimum: Calculus?

How should two nonnegative numbers be chosen so that their sum is 1 and the sum of their squares is (a) as large as possible, (b) as small as possible.

### 3 Answers

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• Anonymous
1 decade ago
Favorite Answer

x + y = 1

y = 1 - x

g = x^2 + (1 - x)^2 = x^2 + 1 - 2x + x^2 = 2x^2 - 2x + 1

g' = 4x - 2 ; 0 = 4x -2 ; 2 = 4x ; x = 1/2 and therefore y = 1/2

so theres a max or min at x = 1/2

g'' = 4, so it is concave up therefore x = 1/2 is a minimum

and because there are no maximum, we know that the highest possible value of g are at its boundaries ( end points )( either x = 0 or x =1). In this case x =0 and x = 1 give the same g so they are both the highest possible value of g

• let the two numbers be x and y

x^2 + y^2 = a where a is a constant

x+y=1

a) x^2 + y^2 =a

2x + 2y(dy/dx) = 1

dy/dx = (1-2x)/2y

when eqn is max or min, dy/dx = 0

(1-2x)/2y = 0

1-2x = 0

x = 0.5

y = 0.5

To find the max value, simply look at the graph of the function

x^2 + y^2 = a

This is the equation for the graph of a circle with center (0,0)

when value is max or min, dy/dx = 0, which means the gradient is 0

The only part of the graph where this happens is on the y-axis ie when x=0 and y = 1 or y=-1. but as your numbers must be positive, y=1

• Anonymous
1 decade ago

Let the two numbers be a and b

then a+b=1

b=1-a

Sume of their squares - S=a^2+b^2

subs in for b

S=a^2+(1-a)^2

S=2a^2-2a+1

dS/da=4a-2

for max and min put dS/da=0

4a-2=0

a=1/2, b=1/2 these values give minimum

maximum is when one is 1 and other is 0.

If you plot the graph f(a)=2a^2-2a+1 you will see that the maximum values are not turning points so will not be calculated by this method.

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