Find two positive real numbers whose product is a maximum and whose sum of the first number and four times the second is 160.
A) $128,8$
B) $88,18$
C) $80,20$
D) $96,16$
E) $56,26$

Question

Quizplus · Accepted Answer

Given the sum of the first number $x$ and four times the second number $y$ is 160, we can write this as $x + 4y = 160$. To maximize the product $xy$, we can use the method of Lagrange multipliers or simply solve for one variable in terms of the other and substitute into the product equation. Solving for $x$ gives $x = 160 - 4y$. The product $xy$ becomes $(160 - 4y)y$. Differentiating this product with respect to $y$ and setting the derivative equal to zero will give the values of $y$ that maximize or minimize the product. Alternatively, recognizing this as a quadratic equation in terms of $y$, we know that the product is maximized when $y$ is at the vertex of the parabola, which occurs at $y = -\frac{b}{2a}$ when the product is in the form $ay^2 + by + c$. Substituting $x = 160 - 4y$ into $xy$ gives a quadratic in $y$, and finding the vertex gives the optimal $y$. Solving for $y$ and then $x$ gives the pair $x = 80, y = 20$, which corresponds to choice C.

Find Two Positive Real Numbers Whose Product Is a Maximum 128,8128,8128,8

Find Two Positive Real Numbers Whose Product Is a Maximum $128,8$