<p>No. This is not in general true.</p>
<p>Maximizing the rectangle formed in the described manner would maximize the product <span class="math-container">$x_1 \times x_2$</span>. In the special case where your utility function takes the form <span class="math-container">$u(x_1,x_2)=x_1x_2$</span>, utility maximizing implies you pick a point on your budget constraint where the rectangle you described is maximized.</p>
<p>Suppose however, that instead your utility function is given by <span class="math-container">$v(x_1,x_2)=x_1^{1/3}x_2^{2/3}$</span> And suppose you are currently maximizing <span class="math-container">$x_1 \times x_2$</span> on your budget constraint. Suppose that <span class="math-container">$p_1=p_2=1$</span>, then maximizing <span class="math-container">$x_1 \times x_2$</span> implies <span class="math-container">$x_1^*=x_2^*$</span>. However maximizing <span class="math-container">$v$</span> with these prices gives: <span class="math-container">$x_1^*=\frac{1}{2}x_2^*$</span>. Thus not maximizing the rectangle subject to your constraint, but still defining a point of tangency between the indifference curve and budget constraint.</p>

CC BY-SA 4.0
2020-09-22T21:36:16.360Z
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
CC BY-SA 4.0
2020-09-22T22:07:42.607Z
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
CC BY-SA 4.0
2020-09-22T22:12:04.557Z
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
CC BY-SA 4.0
2020-09-23T14:13:13.080Z
This was very helpful, thank you.
Drake Groves
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2020-09-23T14:30:03.193Z
you're welcome :)
user18214
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2020-09-22T19:00:21.387Z
2020-09-22T19:25:07.703Z
2020-09-22T19:25:07.703Z
user18214
<p>This is not true in general.</p>
<p>It's even possible that the point of utility maximization <em>minimizes</em> 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 <span class="math-container">$u(x_1,x_2)=(x_1-1)(x_2-1)$</span> and the budget line is <span class="math-container">$4-x_1-x_2=0$</span>. Since both functions are symmetric in <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> we can expect that utility maximization will require <span class="math-container">$x_1=x_2$</span>, and from the budget line this implies <span class="math-container">$x_1=x_2=2$</span>.  At this point:
<span class="math-container">$$u(x_1,x_2)=(2-1)(2-1)=1$$</span>
To confirm that this is a maximum of utility and not a minimum, we may note that the nearby point on the budget line <span class="math-container">$x_1=2.1$</span> and <span class="math-container">$x_2=4-2.1=1.9$</span> implies:</p>
<p><span class="math-container">$$u(x_1,x_2)=(2.1-1)(1.9-1)=(1.1)(0.9)=0.99&lt;1$$</span></p>
<p>The area of the rectangle defined by the point of utility maximisation is <span class="math-container">$x_1x_2=2(2)=4$</span>.  To see that this area is a minimum, out of rectangles defined by points on the indifference curve at which <span class="math-container">$u(x_1,x_2)=1$</span>, suppose first that <span class="math-container">$x_1=3$</span>. To find the corresponding value of <span class="math-container">$x_2$</span> at which <span class="math-container">$u(x_1,x_2)=1$</span> we have:</p>
<p><span class="math-container">$$(3-1)(x_2-1)=1$$</span></p>
<p><span class="math-container">$$x_2=3/2$$</span></p>
<p>and therefore the area of the rectangle is:</p>
<p><span class="math-container">$$x_1x_2=3(3/2)=4.5&gt;4$$</span></p>
<p>Similarly if <span class="math-container">$x_1=4$</span>, we find <span class="math-container">$x_2=4/3$</span> and therefore <span class="math-container">$x_1x_2\approx5.33&gt;4$</span>. Thus the more <span class="math-container">$x_1$</span> is greater than <span class="math-container">$x_2$</span>, the more the area of the rectangle will be greater than <span class="math-container">$4$</span>.  By symmetry the same will apply when <span class="math-container">$x_2&gt;x_1$</span>.</p>

CC BY-SA 4.0
2020-09-22T23:04:41.400Z
2020-09-22T23:04:41.400Z
Adam Bailey
<p>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.</p>

<microeconomics>
Geometric intuition behind the point where the indifference curve meets the budget line
Drake Groves
# Geometric intuition behind the point where the indifference curve meets the budget line

(Asked by: username_0 on 2020-09-22)

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.

**Tags:** microeconomics

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## 2 answers

### Answer by username_1 on 2020-09-22. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_2**: 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.  
> **username_1**: 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 $y_I \geq y_b$  
> **username_1**: 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 $x_I y_I$  
> **username_0**: This was very helpful, thank you.  
> **username_1**: you're welcome :)  

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### Answer by username_2 on 2020-09-22. Score: 0

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 $x\_2>x\_1$.


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