#### Free math classes > Derivative > Examples > Solved exercises > Advanced math level > Solution exercise 1.1

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### Derivatives : solution exercise 1.1 (Advanced math level)

1.1 Optimization : Two numbers are hiding themselves

Two numbers are hiding themselves, however we do know that those two numbers have a sum equal to 12. Find those two numbers :

a) such that the sum of their squares is minimum,

When the sum of their squares, x2 + y2, is minimum we have simultaneously:

(x2 + y2)' = 0 and (x2 + y2)' ' > 0

Reminder:

 How to find the x and y coordinates of an extrema ? Take the first derivative of the function set equal to zero and solve for x. In fact the x-value that makes the first derivative = zero is the x-coordinate of an extrema (minimum or maximum). If the second derivative of a function is positive, then the function is concave up, which means that the curve is curving upwards, like a smile (remember: when you are up, you smile). Hence a minimum implies a positive second derivative. While a negative second derivative implies a concave down curve, which means that the curve is curving downwards like a frowning face (remember: when you are down, you frown). Hence a maximum implies a negative second derivative.

So far we have found the following equations:

(1) x + y = 12

(2) (x2 + y2)' = 0

(3) (x2 + y2)' ' > 0

Those three conditions must be satisfied simultaneously. This means that (1) AND (2) AND (3) must be true.

From (1) we get (1') y = 12 - x. Let us replace y by 12 - x in equation (1) and (2). We get:

(2') (x2 + (12 - x)2)' = 0

(3') (x2 + (12 - x)2)' ' > 0

Let use calculate the derivative of equation (2'):

(x2 + (12 - x)2)' = 0

2x + 2(12-x)(-1) = 0

2x -24 + 2x = 0

24 = 4x

x = 24/4 = 6

However we know thanks to equation (1) that x + y = 12 thus:

y = 12 - x = 12 - 6 = 6

Let us check that the point (x , y) = (6 , 6) satisfies the equation
(3') (x2 + (12 - x)2)' ' > 0:

(x2 + (12 - x)2)' ' > 0

(2x + 2(12-x)(-1))' > 0

2 + 2 > 0

4 > 0

4 is greater than zero, we can then confirm that the second derivative of
(x2 + y2) is positive and thus the function is concave up. (6 , 6) is a minimum x and y coordinates.

Conclusion : 6 and 6 are the numbers we were looking for. Their sum, 6+6 equals 12 and the sum of their squares is minimum.

You can trace the function x2 + (12-x)2 online with our free function grapher. Use a "Range on x-axis" from -10 to 10 and a "Range on y-axis" from -10 to 80.
Trace also the graph of the first derivative thanks to the radio button Derivative. You will then observe that the minimum of the function occurs at x = 6, and that the first derivative is equal to zero at x = 6.

b) such that the product of one number and the square of the other number is maximum,

(1) x + y = 12

(2) (x2 . y)' = 0

(3) (x2 . y)' ' < 0

In fact, the second derivative of equation (3) must be negative in order to have a curve concave down and then an extrema that will be a maximum.

(1') y = 12 - x

(2') (x2 . (12 - x))' = 0

(3') (x2 . (12 - x))' ' < 0

We will figure out the value of x thanks to (2'):

(x2 . (12 - x))' = 0

2x . (12 - x) - x2 = 0

24x - 3x2 = 0

3x( -x + 8) = 0

thus x = 0 or x = 8

Let us check that both values of x found hereabove satisify equation (3'):

(x2 . (12 - x))' ' < 0

(24x - 3x2) ' < 0

(24 - 6x) < 0

24 < 6x

24/6 < x

4 < x

thus x > 4, then the value x = 8 is correct while x = 0 must be discarded because it does not satisfy the equation (3'). Graphically, this means that at x = 0 the function is not concave down but concave up. x = 0 corresponds to a minimum coordinate and what we are looking for is a maximum. This is why we discard that x-value.

Conclusion: x = 8 et y = 12 - x = 12 - 8 = 4. The number we are looking for are 8 and 4.

c) such that the product of one number and the cube of the other number is maximum,

(1) x + y = 12

(2) (x3 . y)' = 0

(3) (x3 . y)' ' < 0

(1') y = 12 - x

(2') (x3 . (12 - x))' = 0

(3') (x3 . (12 - x))' ' < 0

We will find x thanks to equation (2'):

(x3 . (12 - x))' = 0

3x2 . (12 - x) - x3 = 0

36x2 - 4x3 = 0

4x2(9 - x) = 0

thus x = 0 or x = 9

Do the values x = 0 and x = 9 satisfy equation (3') ?

(x3 . (12 - x))' ' < 0

(36x2 - 4x3)' < 0

72x - 12x2 < 0

12 . x . (-x + 6) < 0

If x = 0, then the second derivative 12 . x . (-x + 6) is equal to 0. However it should be smaller then 0. Thus x = 0 must be discarded because it does not satisfy equation (3'). Graphically, x = 0 corresponds to a inflection point or point of inflection (or inflexion), which means that at that point the concavity of the curve is changing. At the inflexion point, the concavity is changing from concave down to concave up and conversely. An inflection point is a point on the curve where the concavity is neither down nor up, there is a transition. However we are looking for a maximum and not for a point of inflexion. We must then discard the value x = 0.

However if x = 9, then the second derivative 12 . x . (-x + 6) is equal to -81 which is smaller than zero and thus satisfies the equation (3').

Conclusion: if x = 9 then y = 12 - x = 12 - 9 = 3. The number we are looking for are 9 and 3.