# Geometric intuition behind the point where the indifference curve meets the budget line

So, this isn't to solve any kind of problem, but rather about the intuition behind the concept. I was wondering if at the optimal point where the indifference curve is tangent to the budget line and utility is maximized, at say, some point A. If you were to draw a vertical line from A down to the horizontal axis, and then draw a horizontal line from A to the vertical axis, would the resulting rectangle have a larger area than any other rectangle created by the other points on the curve? Kind of like the intuition behind least squares regression, but trying to maximize area? Thanks in advance for any help.

No. This is not in general true.

Maximizing the rectangle formed in the described manner would maximize the product $$x_1 \times x_2$$. In the special case where your utility function takes the form $$u(x_1,x_2)=x_1x_2$$, utility maximizing implies you pick a point on your budget constraint where the rectangle you described is maximized.

Suppose however, that instead your utility function is given by $$v(x_1,x_2)=x_1^{1/3}x_2^{2/3}$$ And suppose you are currently maximizing $$x_1 \times x_2$$ on your budget constraint. Suppose that $$p_1=p_2=1$$, then maximizing $$x_1 \times x_2$$ implies $$x_1^*=x_2^*$$. However maximizing $$v$$ with these prices gives: $$x_1^*=\frac{1}{2}x_2^*$$. Thus not maximizing the rectangle subject to your constraint, but still defining a point of tangency between the indifference curve and budget constraint.

• I don't think this quite answers the question which relates to the comparison of rectangles defined not by points on the budget line but by points on the indifference curve. – Adam Bailey Sep 22 '20 at 21:36
• The answer to that question trivially follows from the above discussion. If it is not in general true that the optimal choice maximizes $x_1 x_2$ over all points on the budget line, and we are dealing with the usual textbook case of a convex indifference curve, and we consider the indifference curve that is tangent to the budget line, then for every point on the budget line $x_b,y_b$ there exists a point on the indifference curve $(x_I,y_I)$ such that $x_I=x_b$ and $y_I \geq y_b$ – user18214 Sep 22 '20 at 22:07
• Or in other words: $\forall (x_b,y_b) \exists (x_I,y_I): x_I y_I \geq x_b y_b$ Thus if you are not consuming a bundle that maximizes $x_b y_b$ then you are definitely not consuming a bundle that maximizes $x_I y_I$ – user18214 Sep 22 '20 at 22:12
• This was very helpful, thank you. – Drake Groves Sep 23 '20 at 14:13
• you're welcome :) – user18214 Sep 23 '20 at 14:30

This is not true in general.

It's even possible that the point of utility maximization minimizes the area of the rectangle, out of the rectangles defined by points on the indifference curve which is tangential to the budget line. Suppose for example that the utility function is $$u(x_1,x_2)=(x_1-1)(x_2-1)$$ and the budget line is $$4-x_1-x_2=0$$. Since both functions are symmetric in $$x_1$$ and $$x_2$$ we can expect that utility maximization will require $$x_1=x_2$$, and from the budget line this implies $$x_1=x_2=2$$. At this point: $$u(x_1,x_2)=(2-1)(2-1)=1$$ To confirm that this is a maximum of utility and not a minimum, we may note that the nearby point on the budget line $$x_1=2.1$$ and $$x_2=4-2.1=1.9$$ implies:

$$u(x_1,x_2)=(2.1-1)(1.9-1)=(1.1)(0.9)=0.99<1$$

The area of the rectangle defined by the point of utility maximisation is $$x_1x_2=2(2)=4$$. To see that this area is a minimum, out of rectangles defined by points on the indifference curve at which $$u(x_1,x_2)=1$$, suppose first that $$x_1=3$$. To find the corresponding value of $$x_2$$ at which $$u(x_1,x_2)=1$$ we have:

$$(3-1)(x_2-1)=1$$

$$x_2=3/2$$

and therefore the area of the rectangle is:

$$x_1x_2=3(3/2)=4.5>4$$

Similarly if $$x_1=4$$, we find $$x_2=4/3$$ and therefore $$x_1x_2\approx5.33>4$$. Thus the more $$x_1$$ is greater than $$x_2$$, the more the area of the rectangle will be greater than $$4$$. By symmetry the same will apply when $$x_2>x_1$$.