Chapter 2.1, Problem 58E

To determine

To find:The twonumbers such that the sum of the first and three times the second is 42, and their product is maximum.

Answer to Problem 58E

The product of two numbers such that the sum of the first and three times the second is 42, is maximum when they are21 and 7.

Explanation of Solution

Given data:

The sum of the first and three times the second is 42.

Concept used:

The vertex of downwards opening quadratic function is maxima of the function. The standard form of the quadratic function is $y=a{\left(x-h\right)}^2+k$ that has vertex $\left(h,k\right)$ . If $a$ is negative then parabola is concave downwards,

Calculations:

Let x and z be two numbers such that the sum of x and 3z is 42 and the product $xz$ is maximum. We denote the product of numbers by y. Thus,

  $x+3z=42$...............(1)

  $y=xz$...............(2)

From results (1) and (2), we have

  $y=z\left(42-3z\right)=-3z^2+42z$

  $y=-3\left(z^2-14z\right)$...............(3)

Now we obtain standard form of quadratic function $y=-3\left(z^2-14z\right)$ and find its vertex.

  $y=-3\left(z^2-14z+49\right)+147$

  $y=-3{\left(x-7\right)}^2+147$...............(4)

Since, $a=-3<0$ the function $y=-3{\left(x-7\right)}^2+147$ is a downwards opening parabola with vertex $\left(7,147\right)$ and its maximum occurs at the vertex $\left(z\text{,}{y}_{\text{maximum}}\right)=\left(7,147\right)$ hat implies $y_{\text{maximum}}=147$ for $z=6$ then $x=\frac{147}{7}=21$

This shows, the product $y=xz$ of two numbers x, zis maximum when $x=21$ and other number $z=7$ .

Thus, the product of two numbers such that the sum of the first and three times the second is 42, is maximum when they are21 and 7.

Conclusion:

The product of two numbers such that the sum of the first and three times the second is 42, is maximum when they are21 and 7.