# Finding area of largest rectangle between the axes and a line

The question is as follows: Find the area of the largest rectangle that has sides parallel to the coordinate axes, one corner at the origin and the opposite corner on the line 3x+2y=12 in the first quadrant.

I get that the equation I have to maximize is in the form of A=bh but I don't know how to eliminate one of the variables to continue.

• Hint: If the opposite corner has coordinates $(h,k)$, then $3h+2k=12$ and the area is given by $A=hk$. Oct 8, 2018 at 19:10

Since the bottom left corner of the rectangle is at the origin, then the $$(x,\,y)$$ coordinates of the top right corner will be the base and height (draw a figure to help visualize). We know that this point is on the line $$3x+2y=12$$, so that $$y=-\frac{3}{2}x+6$$. Plugging this in gives $$A=x(-\frac{3}{2}x+6)$$, which has only one variable.

Suppose that your rectangle has vertices $$(0, 0)$$, $$(x, 0)$$, $$(0, y)$$, and $$(x, y)$$, where $$x > 0$$, $$y > 0$$, and $$3x + 2y = 12. \tag{1}$$ Then the area of your rectangle is given by $$A = xy. \tag{2}$$ But (1) implies that $$y = 6 - \frac{3x}{2}. \tag{3}$$ Putting the value of $$y$$ from (3) into the formula in (2), we obtain $$A = A(x) = x \left( 6 - \frac{3x}{2} \right) = 6x - \frac{3x^2}{2}. \tag{4}$$ Now (4) gives area $$A$$ as a function of $$x$$ for $$x > 0$$.

Differentiating both sides of (4) w.r.t. $$x$$ we obtain $$A^\prime(x) = 6 - 3x.$$ Thus we see that $$A^\prime(x) \ \begin{cases} > 0 \ & \ \mbox{ for } x < 2, \\ = 0 \ & \ \mbox{ for } x = 2, \\ < 0 \ & \ \mbox{ for } x > 2. \end{cases}$$ Thus the area attains its (relative) maximum value at $$x = 2$$, and since this is the only relative extreme value of $$A$$, this is in fact the absolute maximum value of $$A$$.

Therefore the largest possible area is given by $$A(2) = 12 - 6 = 6.$$

Any point on the line can be represented as (x,(12-3x)/2)

Now we have to find the maximum area of a rectangle given one point as (0,0) and opposite point as $$(x,(12-3x)/2$$)

Area=$$(x)*(12-3x)/2$$

We have to maximize this area taking derivative and equating to $$0$$ we get value of $$x=2$$

Substituting this value in area we get area=6 which should be maximum