Solving Math Problems: Equations, Matrices, and Optimization
Problem 1: Age Calculation
Three men’s ages total 85 years. Within 5 years, what will be the sum of their ages? They now have two children. Within two years, the middle child’s age will be half the sum of the current ages of the youngest and oldest children. How old is everyone?
Solution:
Let:
Problem 2: Matrix Equation
Determine the matrix A that satisfies the equation A – 2B = AB’, where B = 
and B’ represents the transpose of B.
Solution:
We need to solve: A – 2
= A
That is:
A – A
= 
A
–
= A
= 
Therefore, A =
=
= 
Problem 3: System of Equations
Solution:
The system can be written in matrix form:
= 
Rank(A) -> 
= -14
Rank(A) = 3 = Rank(AB)
This is a compatible and determined system. The solution is:
x = 
= 
y = 
= 
z = 
= 0
Problem 4: Storage Optimization
A store stocks sunflower oil and olive oil. To meet customer demand, they must have a minimum of 29 cans of sunflower oil and 40 drums of olive oil. Additionally, the number of cans of olive oil should not be less than half the number of cans of sunflower oil. The total storage capacity is 150 cans. Knowing that the storage cost of an olive oil drum is 10 euros and a can of sunflower oil is 5 euros, how many cans of each type should be stored to minimize expenditure? What is that minimum expenditure?
Solution:
Let:
We aim to maximize or minimize the objective function f(x, y) = 10x + 5y
subject to the restrictions:
The objective function value at these points is:
f(A) = 490
f(B) = 580
f(C) = 1000
f(D) = 1300
Therefore, to minimize expenditure, we need to store 29 cans of sunflower oil and 40 cans of olive oil. To maximize expenditure, we need to store 110 cans of sunflower oil and 40 cans of olive oil.
More Math Challenges
Problem 1: Real Estate Sales
A real estate agent has sold a total of 65 parking spaces in 3 different neighborhoods. Proceeds from the sale of a parking space in residential area A is 2,000 euros, 4,000 euros in residential area B, and 6,000 euros in residential area C.
We know that 50% more spaces were sold in area B than in area C. Calculate the number of parking spaces sold in each development, knowing that the profit from those sold in area C is equal to the sum of the profits from those sold in areas A and B.
Solution:
Let:
Before solving the system, we simplify the second equation:
Problem 2: Inverse Matrix and Equation
Given the matrix A = 
a) Find its inverse.
b) Solve the equation XA 
+ 5A = 
Solution:
a) A
= 
= 
b) X
=
– 5
X
=
– 5
X = 
X = 
Problem 3: Inequations and Optimization
a) Graph the solution set of the system of inequations:
b) Determine the vertices of the region obtained in the previous paragraph.
c) Calculate the point where the function f(x, y) = 3x – y reaches its minimum and find the minimum value.
Solution:
b) A = 
-> A(1, 1)
B = 
-> B(2, 
)
C = 
-> C(4, -1)
D = 
-> D(3, -2)
c) f(A) = 2
f(B) = 4.5
f(C) = 13
f(D) = 11
The minimum is reached at point A(1, 1) with a value of 2.
Problem 4: Company Profit Analysis
The income statement (profit or loss) in millions of euros of a company is given by this function of x years of its existence:
Y = 
a) Since what year has the company started experiencing losses?
b) When does the company reach its maximum payout? How much is it?
Solution:
a) The company has losses when the function is negative. Since x represents the years of existence of the business, it cannot be negative. Therefore, the function will indicate a loss when the numerator is negative, that is, when 5
+ 20x – 25 
, which occurs after the first year (x = 1).
b) 
This equals zero when the numerator is zero, that is:
Since the function is increasing to the left of 7 and decreasing to the right, at x = 7 there is a maximum, and the maximum profit will be:
Problem 5: Function Analysis
a) Calculate the absolute maximum and minimum of the function f(x) = 
in the interval [1, 4]. Justify that the points found are absolute maximum or minimum.
b) Study the continuity in the interval [0, 4] of this function:
f(x) = 
Solution:
a) f'(x) = 3x
– 12x + 9, which equals zero for 
At x = 3 there is a minimum.
f(1) = f(4) = 5
At x = 4 there is also a maximum.
b) The function is continuous within the intervals. At the point of discontinuity:
Therefore, the function is always continuous on the interval [0, 4].