# Finding the maximum area of a quadrilateral when three points are given

I am working on problems in the chapter "Applications of Derivatives". I encountered the following problem:

Question:

Four points A,B,C, and D lie in that order on the parabola $$y=ax^2+bx+c$$. The coordinates of A,B, and D are $$(-2,3)$$,$$(-1,1)$$, and $$(2,7)$$ respectively. The coordinates of C for which the area of the quadrilateral ABCL is maximum is:

(A) $$(1/2,7/4)$$

(B) $$(1/2,-7/4)$$

(C) $$(-1/2,7/4)$$

(D) $$(-1/2,-7/4)$$

My Approach:

Introduction:

I found the equation of the parabola using the given coordinates. Then computed the area of the quadrilateral by considering an arbitrary value for point C. Then I found the coordinates of C by maximizing the value of A by using the concept of maxima. What I am looking for is an alternate proof or a property which relates the coordinates of third vertex of a quadrilateral when three points are fixed under certain constraints, to maximize the area.

We have been given the coordinates A$$(-2,3)$$,B$$(-1,1)$$ and D$$(2,7)$$.

Since these points lie on the given parabola $$y=ax^2+bx+c$$, on substituting the coordinates of A,B and D we get the following linear equations in variables $$a,b,$$ and $$c$$:

$$4a-2b+c=3$$

$$a-b+c=1$$

$$4a+2b+c=7$$

On solving these three equations, I got $$a=1,b=1,$$ and $$c=1$$. So the equation of the parabola is $$y=x^2+x+1$$.

In the following graph, the points A,B and D are fixed, whereas point C is a variable point on the parabola. We need to find the coordinates of point C such that the area of the quadrilateral is maximum.

In coordinate geometry, I came across the following formula to compute the area of a $$n$$ sided polygon when the coordinates are given:

Area A $$= \frac 1 2 \left( \left| {\begin{array}{cc}x_1 & x_2 \\y_1 & y_2 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_2 & x_3 \\y_2 & y_3 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_3 & x_4 \\y_3 & y_4 \\ \end{array} } \right|+ \dots +\left| {\begin{array}{cc}x_{n-1} & x_n \\y_{n-1} & y_n \\ \end{array} } \right|+\left| {\begin{array}{cc}x_n & x_1 \\y_n & y_1 \\ \end{array} } \right| \right)$$

where $$(x_1,y_1),(x_2,y_2),(x_3,y_3),\dots ,(x_{n-1},y_{n-1}),(x_n,y_n)$$ are coordinates of the $$n$$ vertices of the $$n$$ sided polygon taken in order. Here |.| denotes determinant of the matrix.

Now in the given question we are supposed to find the maximum area of a quadrilateral (4 sided polygon). So using the above formula, we get the following:

Area A $$= \frac 1 2 \left( \left| {\begin{array}{cc}x_1 & x_2 \\y_1 & y_2 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_2 & x_3 \\y_2 & y_3 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_3 & x_4 \\y_3 & y_4 \\ \end{array} } \right|+\left| {\begin{array}{cc}x_4 & x_1 \\y_4 & y_1 \\ \end{array} } \right| \right)$$

where where $$(x_1,y_1),(x_2,y_2),(x_3,y_3),$$ and $$(x_4,y_4)$$ are the coordinates of points A,B,C, and D respectively.

Now let the $$x$$-coordinate of point C be $$h$$. So its $$y$$-coordinate (from the equation of parabola) is $$h^2+h+1$$. So coordinates of point C be represented as $$(h^2+h+1)$$.

After substituting the values of the coordinates in the equation for area, I obtained the following equation:

Area A$$=\frac 1 2 -3h^2+3h+18$$

Area A attains the maximum value at $$h=1/2$$.

Hence the coordinates of point C is $$(1/2,7/4)$$ - option (A). My answer is correct.

Doubt:

Is there any other formal way of solving this problem? Is there any property for choosing the fourth vertex of the quadrilateral when three vertices are given, in order to get the maximum area under given constraints?

Credits to @Siddhant. After reading his answer, another method came to my mind and I wish to share it here.

The coordinates of points A,B and D have been given in the question. So these points are fixed. Point C is the variable point on the given parabola. We need to choose the appropriate value of C so that the area of the quadrilateral ABCD is maximum.

Let us consider the following diagram.

Area of the quadrilateral ABCD is nothing but the sum of areas of the triangles ABD and BCD. Since points A,B and D are fixed, the area of triangle ABD is constant for the given question. Since C is a variable point and is not moving parallel to the line BD, the area of triangle BCD is not constant.

Our objective is to maximize the area of the quadrilateral ABCD. Since area of ABD is fixed. We need to maximise the are of triangle BCD.

We know that, area of a triangle is given by $$\frac 1 2 bh$$ where $$b$$ and $$h$$ are the base and hypotenuse of the triangle respectively. In triangle BCD, let us consider the base as the line segment BD and the height represented by the green coloured line in the above diagram. Since B and D are fixed, the length of base BD is fixed. Now, in order to maximise the area of triangle BCD, it is sufficient to find the coordinates for the point C when it is at the maximum distance from the line segment BD(base of BCD).

Now let us find the equation of the line BD using the coordinates of B and D.

$$\frac{y-1}{x+1}=\frac 6 3 = 2$$ $$y-1=2(x+1)$$ $$2x-y+3=0$$

Next, let us compute the distance (or height $$h$$) of the variable point C$$(p,p^2+p+1)$$ from the base BD as shown below:

$$h=\frac{2p-(p^2+p+1)+3}{\sqrt{2^2+1^2}}$$ $$h=\frac{-p^2+p+2}{\sqrt 3}$$

We have obtained $$h$$ as a function of $$p$$. Now we need to maximize $$h$$. It is easy to see that the quadratic in $$p$$ attains the highest value at $$p=1/2$$.

Hence the coordinate for point C such that the area of quadrilateral ABCD is maximum is $$(1/2,7/4)$$.

• Nice geometric interpretation. Commented Oct 7, 2019 at 11:52
• Alternatively, you can require that the tangent to the parabola in $p$ is parallel to BD.
– chi
Commented Oct 7, 2019 at 18:07

Its not a completely different approach but just a way to reduce the matrix computations.

$$1^{st}$$ step - Get the parabola $$y(x) = x^2 + x + 1$$ which you got it with the points A,B,D.

Now, if you join BD, the total area you interested in maximizing would be $$Area(ABD)+Area(BCD)$$ triangles. As area cannot be negative, you need to maximize just Area(BCD) as points A,B,D are known and so is the value of Area(ABD).

Therefore computing the $$Area(BCD)$$ you can directly use the area of triangle formula =$$(1/2)det(A)$$ where $$A =$$ $$\left[ \begin{matrix} h & k & 1\\ -1 & 1 & 1 \\ 2 & 7 & 1\\ \end{matrix}\right]$$

This is a simple determinant which we get as $$Area(BCD) = (1/2)(3k-6h-9)$$. We need to maximize this following the constraint $$k=h^2 + h + 1$$. Just substitute for $$k$$ and differentiate with respect to $$h$$, we get $$h=1/2$$, and so $$k=7/4$$.

There is also a faster way of finding the value of coordinates of point $$C$$. After finding the equation of the curve. Just utilize the options, substitute each value of $$C$$. The one which satisfies is the correct answer. Luckily here only one option matches - option $$A$$. This may help you in exams where time counts!

• I agree your method is the simplest. Here it works great. But, I was seeing for method that work in all conditions. That's why I asked for a formal method and not just a method. Commented Oct 7, 2019 at 11:28

Substitute points $$( A,B,D )$$ into $$ax^2 + bx + c,$$ and solve using Cramer's Rule of three simultaneous equations in order to evaluate $$( a,b,c)=(1,1,1)$$.

The point $$(x_C,y_C)$$ can be found by drawing a tangent by differentiating $$y = ax^2+bx+c$$ w.r.t $$x$$ at future point C for maximum area.

This is so because maximum height of triangle $$(Area = base DB \cdot h_{max})$$ is between parallel lines.

Slope of DB =

$$2 ax_C +b= \dfrac{6}{3}=2 \rightarrow x_C=\dfrac12;$$

$$y_C= (\dfrac12)^2 + 1\cdot \, \frac12 + 1= \frac74.$$