Question

One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes that can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

Answer 1

Hint – In this problem let x be the number of the cakes that can be formed of one kind and y be the number of cakes that can be formed of other kinds. Construct a linear equation resembling inequality using the relation given in the problem. Since we are concerned about the intersection of these inequality thus two linear equations involving two variables will be formed. Solve them to get the required answer.
Complete step-by-step solution -
Let x be the number of cakes of one kind.
And y be the number of cakes of other kinds.
Now according to the question we have to maximize (x + y).
Now it is given that 200 gm of flour and 25 gm of fat is used to make one kind of cake.
And 100 gm of flour and 50 gm of fat is used to make other kinds of cake.
Now it is given that there are a total 5 kg of flour and 1 kg of fat.
Now as we know that 1 kg = 1000 gm.
So 5 kg = 5000 gm.
So there are 5000 gm of flour and 1000 gm of fat.
Now construct the linear equation according to above information we have,
$\Rightarrow 200x + 100y \leqslant 5000$..................... (1)
And
$\Rightarrow 25x + 50y \leqslant 1000$........................ (2)
Now in equation (1) and (2) divide by 100 and 25 respectively we have,
$\Rightarrow 2x + y \leqslant 50$
And, $x + 2y \leqslant 40$
Now we have to make both kinds of cakes so the point of intersection is the required solution.
$\Rightarrow 2x + y = 50$........................ (3)
And, $x + 2y = 40$..................... (4)
So from equation (3) we have,
$\Rightarrow y = 50 - 2x$
Now substitute this value in equation (4) we have,
$\Rightarrow x + 2\left( {50 - 2x} \right) = 40$
Now simplify this equation we have,
$\Rightarrow x + 100 - 4x = 40$
$\Rightarrow 3x = 100 - 40 = 60$
$\Rightarrow x = \dfrac{{60}}{3} = 20$
Now substitute this value in equation (3) we have,
$\Rightarrow 2\left( {20} \right) + y = 50$
$\Rightarrow y = 50 - 40 = 10$
Therefore maximum number of cakes which can be made are
$\Rightarrow \left( {x + y} \right) = 20 + 10 = 30$
So we can make a maximum of 30 cakes from 5kg of flour and 1 kg of fat.
So this is the required answer.

Note – In this problem the two linear equations in two variables are being solved using the method of substitution that is one variable is taken in terms of another variable and then substituted into another equation to get the variables. There can be another method to solve them; it is called the method of elimination, in which one variable of the two equations is eliminated by making the coefficients same for that variable in both the equations. Then this variable is eliminated by performing simple algebraic operations.