Find the area of the largest rectangle that can be drawn in the first quadrant with one corner at the origin, and another corner on the line 3x + y = 1.

Answer 1

We know one corner of the triangle must be at (0,0), the origin. We know another corner must be at a point along 3x+y=1 in the first quadrant. From this, we know the bounds of the box will be from 0 to some value on the line. We know the surface area of a rectangle is defined based upon the length of the two sides multiplied. We can represent this as x*y=a, where x and y are the values in the line equation. Isolating y, you get y=1-3x. We can substitute this into our area equation to get x*(1-3x)=a. This will give us the surface area of a rectangle where one point is on the origin and another is on the line. Reducing this, you get a=x-3x² . You can then use your preferred method of optimizing this equation to find the maximum of this quadratic equation. I will go with a first derivative test because this is a quadratic equation with one extremum we know will be a maximum because this function is concave down. If we were unsure of this, we would do a second derivative test or check the points around our test point.

Taking the first derivative using the power rule, you get a'=1-6x

setting this equal to 0, we get

0=1-6x

1=6x

1/6=x

This tells us the extrema of our area function is at x=1/6. Because that is the absolute maximum due to the function for the area being quadratic, you know that is the largest surface area possible for the conditions we create the equation for, where one point is the origin and another is on the line. Bringing back our area equation (a=x-3x³), we can plug in our found x value and get 1/6-3*(1/6)²=1/12. Thus we now know the largest surface area possible of this rectangle is 1/12.

Answer 2

y = -3x + 1

A = xy

A = x(-3x + 1)

A = -3x² + x

Maximum at x = -b / 2a, or x = -1 / 2(-3).