In the standard (x,y) coordinate plane, a region in the first quadrant is bounded by the x… — ACT

In the standard $(x,y)$ coordinate plane, a region in the first quadrant is bounded by the $x$ -axis, the $y$ -axis, and the line with equation $ax + by = c$ , where $a$ , $b$ , and $c$ are positive constants. The line passes through the point $(8, 18)$ . If the area of this region is minimized when $a = 3$ , what is the value of $c$ ?

Answer: 48

Answer

Substituting the given point and

a = 3

into the equation yields

c = 24 + 18b

. The area of the triangle formed by the intercepts is

A = \frac{c^2}{6b}

. Substituting

c

gives

A = 6(9b + 24 + \frac{16}{b})

. Using AM-GM, the minimum occurs when

9b = \frac{16}{b}

, resulting in

b = \frac{4}{3}

. Using this value, we find

c = 48

Step-by-Step Solution

Substitute the point

(8, 18)

and

a = 3

into the equation

ax + by = c

24 + 18b = c

This establishes a relationship between the constants

b

and

c

based on the given point that lies on the line.

Calculate the

x

-intercept and

y

-intercept of the line.

x

-intercept is at

x = \frac{c}{3}

, and

y

-intercept is at

y = \frac{c}{b}

The boundary of the region in the first quadrant is defined by these coordinate intercepts.

Formulate the area

A

of the right triangle bounded by the axes and the line.

A = \frac{c^2}{6b}

The area of a right triangle with vertices at the origin and the intercepts is

\frac{1}{2} \cdot \text{base} \cdot \text{height}

Substitute

c = 24 + 18b

into the area formula and simplify.

A = 6\left(9b + 24 + \frac{16}{b}\right)

Expressing the area as a single-variable function of

b

allows us to find its minimum value.

Apply the AM-GM inequality to minimize the variable term

9b + \frac{16}{b}

b = \frac{4}{3}

minimizes the expression.

The sum of two positive terms is minimized when the terms are equal, so

9b = \frac{16}{b} \implies b^2 = \frac{16}{9} \implies b = \frac{4}{3}

Calculate the value of

c

using the minimizing value of

b

c = 48

Substituting

b = \frac{4}{3}

into the relation

c = 24 + 18b

yields the constant value

c

for the minimum area.

Key Concept

Minimizing the area bounded by a line and the coordinate axes using linear equation forms and algebraic minimization.

Alternative Method

Instead of using the AM-GM inequality, you can find the minimum by taking the derivative of the area function

A(b) = 54b + 144 + \frac{96}{b}

with respect to

b

. Setting the derivative

A'(b) = 54 - \frac{96}{b^2} = 0

yields

b^2 = \frac{96}{54} = \frac{16}{9}

, which gives

b = \frac{4}{3}

for

b > 0

Estimated Time:3m 0s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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