Farmer Jane has many hens that have laid many eggs. Before the eggs hatch, Jane wants to count how many eggs are blue, how many are brown, and how many are white.
Jane counts that, of all the eggs, 6 are blue and 25% are brown.
If 10% of the eggs are blue, how many of the eggs are white?
a) 15
b) 21
c) 39
d) 65

The correct option is;

c) 39

Step-by-step explanation:

The given information are;

The number of eggs that are blue = 6

The percentage of eggs that are brown = 25%

The percentage of eggs that are blue = 10%

Let the number of eggs Jane has = X

Therefore;

10% of X = 10/100 × X = 6

X = 6 × 100/10 = 60

The number of eggs Jane has = X = 60

The percentage of the eggs that are white eggs = 100 - (The percentage of eggs that are brown + The percentage of eggs that are blue)

The percentage of the eggs that are white eggs = 100 - (10 + 25) = 65%

The percentage of the eggs that are white eggs = 65%

∴ The number of  eggs that are white = 65% of X = 65/100 × X= 65/100× 60 = 39 eggs

The number of  eggs that are white = 39 eggs.

x = 6, y =3, and the minimum = 45

Step-by-step explanation:

There are plenty of ways to solve this problem.

The Lagranian multiplier method is chosen (it is simple to apply in this case).

Given: A = x^2 + y^2

Find: x and y that satisfy 4x + 2y = 30 or 4x + 2y - 30 = 0 and A is minimum.

This is the same as finding x, y, and lambda that make the following L minimum.

L= x^2 + y^2 + lambda(4x + 2y - 30)

To find x, y, and lambda, we take the derivative of L with respect to them, set the derivative to 0, and solve the equations.

L'(x) = 2x + 4(lambda) = 0    (1)

L'(y) = 2y + 2(lambda) = 0    (2)

L'(lambda) = 4x + 2y - 30 = 0   (3)

Multiply (2) by 2 and subtract to (1), we have:

4y - 2x = 0  (*)

Combine (*) with (3) we have:

4y - 2x = 0

4x + 2y - 30 = 0

Multiply the 1st equation by 2 and add to the 2nd equation:

=>  10y - 30 = 0

=> y = 3

=> x = 4y/2 = 4(3)/2 = 6 (as (*))

=> lambda = -y = -3  (as (2))

Substitute x = 6, y = 3, and lambda = -3 into L, or substitute x = 6 and y = 3 into A to get the minimum.

Both ways would lead to the same answer.

Check:

L = x^2 + y^2 + lambda(4x + 2y - 30)

  = 6^2 + 3^2 + (-3)[4(6) + 2(3) - 30]

  = 36 + 9 + (-3)(0)

  = 45

A = x^2 + y^2

   = 36 + 9

   = 45

Hope this helps!