Express the distance two points and optimize the area of a triangle.
$begingroup$
Consider the parabola $mathcal{P}$ of equation $;y = x^2,$ and the line $L$ of equation $y = x+6.$ Let $P(x_P,y_P)$ be a point on the arc of the parabola $mathcal{P}$ below $L.$ Let $A$ and $B$ be the points of intersection of $mathcal{P}$ and $L.$
a) Find an equation of the line $L_P$ passing through $P$ which is perpendicular to $L.$
b) Let $Q_P$ be the point of intersection of $L$ and $L_P.$ Express the distance between $P$ and $Q_P$ using only $x_P$ and $y_P.$
c) Find the point $P$ on the arc of $mathcal{P}$ below $L$ such that the area of the triangle $APB$ is maximal.
I had no problem doing part a) and got $y=8-x,$ however, I cannot continue further with part b) and c) 
without hesitation. Please help me!
derivatives optimization area
edited Nov 30 '18 at 5:03
user376343
3,3582826
asked Nov 30 '18 at 4:05
Josh TealJosh Teal
588
$endgroup$
add a comment |
$begingroup$
Consider the parabola $mathcal{P}$ of equation $;y = x^2,$ and the line $L$ of equation $y = x+6.$ Let $P(x_P,y_P)$ be a point on the arc of the parabola $mathcal{P}$ below $L.$ Let $A$ and $B$ be the points of intersection of $mathcal{P}$ and $L.$
a) Find an equation of the line $L_P$ passing through $P$ which is perpendicular to $L.$
b) Let $Q_P$ be the point of intersection of $L$ and $L_P.$ Express the distance between $P$ and $Q_P$ using only $x_P$ and $y_P.$
c) Find the point $P$ on the arc of $mathcal{P}$ below $L$ such that the area of the triangle $APB$ is maximal.
I had no problem doing part a) and got $y=8-x,$ however, I cannot continue further with part b) and c) without hesitation. Please help me!
derivatives optimization area
edited Nov 30 '18 at 5:03
user376343
3,3582826
asked Nov 30 '18 at 4:05
Josh TealJosh Teal
588
$endgroup$
$begingroup$
There are many points on the parabola, bellow L. From your answer to a) it is clear you have chosen one particular point. That's why your coefficient 8 is fixed and you cannot continue. The point P has to be arbitrary. In the equation of $L_p$ should be a parameter.
$endgroup$
– user376343
Nov 30 '18 at 5:18
add a comment |
$begingroup$
Consider the parabola $mathcal{P}$ of equation $;y = x^2,$ and the line $L$ of equation $y = x+6.$ Let $P(x_P,y_P)$ be a point on the arc of the parabola $mathcal{P}$ below $L.$ Let $A$ and $B$ be the points of intersection of $mathcal{P}$ and $L.$
a) Find an equation of the line $L_P$ passing through $P$ which is perpendicular to $L.$
b) Let $Q_P$ be the point of intersection of $L$ and $L_P.$ Express the distance between $P$ and $Q_P$ using only $x_P$ and $y_P.$
c) Find the point $P$ on the arc of $mathcal{P}$ below $L$ such that the area of the triangle $APB$ is maximal.
I had no problem doing part a) and got $y=8-x,$ however, I cannot continue further with part b) and c) without hesitation. Please help me!
derivatives optimization area
edited Nov 30 '18 at 5:03
user376343
3,3582826
asked Nov 30 '18 at 4:05
Josh TealJosh Teal
588
$endgroup$
Consider the parabola $mathcal{P}$ of equation $;y = x^2,$ and the line $L$ of equation $y = x+6.$ Let $P(x_P,y_P)$ be a point on the arc of the parabola $mathcal{P}$ below $L.$ Let $A$ and $B$ be the points of intersection of $mathcal{P}$ and $L.$
a) Find an equation of the line $L_P$ passing through $P$ which is perpendicular to $L.$
b) Let $Q_P$ be the point of intersection of $L$ and $L_P.$ Express the distance between $P$ and $Q_P$ using only $x_P$ and $y_P.$
c) Find the point $P$ on the arc of $mathcal{P}$ below $L$ such that the area of the triangle $APB$ is maximal.
I had no problem doing part a) and got $y=8-x,$ however, I cannot continue further with part b) and c) without hesitation. Please help me!
derivatives optimization area
derivatives optimization area
edited Nov 30 '18 at 5:03
user376343
3,3582826
asked Nov 30 '18 at 4:05
Josh TealJosh Teal
588
edited Nov 30 '18 at 5:03
user376343
3,3582826
asked Nov 30 '18 at 4:05
Josh TealJosh Teal
588
edited Nov 30 '18 at 5:03
user376343
3,3582826
edited Nov 30 '18 at 5:03
user376343
3,3582826
edited Nov 30 '18 at 5:03
user376343
3,3582826
3,3582826
asked Nov 30 '18 at 4:05
Josh TealJosh Teal
588
asked Nov 30 '18 at 4:05
Josh TealJosh Teal
588
asked Nov 30 '18 at 4:05
Josh TealJosh Teal
588
588
$begingroup$
There are many points on the parabola, bellow L. From your answer to a) it is clear you have chosen one particular point. That's why your coefficient 8 is fixed and you cannot continue. The point P has to be arbitrary. In the equation of $L_p$ should be a parameter.
$endgroup$
– user376343
Nov 30 '18 at 5:18
add a comment |
$begingroup$
There are many points on the parabola, bellow L. From your answer to a) it is clear you have chosen one particular point. That's why your coefficient 8 is fixed and you cannot continue. The point P has to be arbitrary. In the equation of $L_p$ should be a parameter.
$endgroup$
– user376343
Nov 30 '18 at 5:18
$begingroup$
There are many points on the parabola, bellow L. From your answer to a) it is clear you have chosen one particular point. That's why your coefficient 8 is fixed and you cannot continue. The point P has to be arbitrary. In the equation of $L_p$ should be a parameter.
$endgroup$
– user376343
Nov 30 '18 at 5:18
$begingroup$
There are many points on the parabola, bellow L. From your answer to a) it is clear you have chosen one particular point. That's why your coefficient 8 is fixed and you cannot continue. The point P has to be arbitrary. In the equation of $L_p$ should be a parameter.
$endgroup$
– user376343
Nov 30 '18 at 5:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: You can easily find out the coordinates of $Q_P$ by solving the system of linear equations
begin{align}
y-x=6\
x+y=8
end{align}
Knowing $Q_P$ you can express the distance between $P$ and $Q_P$ in terms of $(x_P,y_P)$.
For the third part, first find out the distance between $A$ and $B$. Let that distance be $b$. Let the distance between $P$ and $Q_P$ be $h$. Then the area of triangle is $frac{bh}{2}$. Note that the only variables are $x_P$ and $y_P$. Now use the relation $y=x^2$ and take the derivative of area with respect to either $x$ or $y$. (Depends on your substitution.) I hope you know how to proceed.
P.S:My answer is based on the assumption that your first part is correct.
answered Nov 30 '18 at 4:39
Thomas ShelbyThomas Shelby
2,250220
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Hint: You can easily find out the coordinates of $Q_P$ by solving the system of linear equations
begin{align}
y-x=6\
x+y=8
end{align}
Knowing $Q_P$ you can express the distance between $P$ and $Q_P$ in terms of $(x_P,y_P)$.
For the third part, first find out the distance between $A$ and $B$. Let that distance be $b$. Let the distance between $P$ and $Q_P$ be $h$. Then the area of triangle is $frac{bh}{2}$. Note that the only variables are $x_P$ and $y_P$. Now use the relation $y=x^2$ and take the derivative of area with respect to either $x$ or $y$. (Depends on your substitution.) I hope you know how to proceed.
P.S:My answer is based on the assumption that your first part is correct.
answered Nov 30 '18 at 4:39
Thomas ShelbyThomas Shelby
2,250220
$endgroup$
add a comment |
$begingroup$
Hint: You can easily find out the coordinates of $Q_P$ by solving the system of linear equations
begin{align}
y-x=6\
x+y=8
end{align}
Knowing $Q_P$ you can express the distance between $P$ and $Q_P$ in terms of $(x_P,y_P)$.
For the third part, first find out the distance between $A$ and $B$. Let that distance be $b$. Let the distance between $P$ and $Q_P$ be $h$. Then the area of triangle is $frac{bh}{2}$. Note that the only variables are $x_P$ and $y_P$. Now use the relation $y=x^2$ and take the derivative of area with respect to either $x$ or $y$. (Depends on your substitution.) I hope you know how to proceed.
P.S:My answer is based on the assumption that your first part is correct.
answered Nov 30 '18 at 4:39
Thomas ShelbyThomas Shelby
2,250220
$endgroup$
add a comment |
$begingroup$
Hint: You can easily find out the coordinates of $Q_P$ by solving the system of linear equations
begin{align}
y-x=6\
x+y=8
end{align}
Knowing $Q_P$ you can express the distance between $P$ and $Q_P$ in terms of $(x_P,y_P)$.
For the third part, first find out the distance between $A$ and $B$. Let that distance be $b$. Let the distance between $P$ and $Q_P$ be $h$. Then the area of triangle is $frac{bh}{2}$. Note that the only variables are $x_P$ and $y_P$. Now use the relation $y=x^2$ and take the derivative of area with respect to either $x$ or $y$. (Depends on your substitution.) I hope you know how to proceed.
P.S:My answer is based on the assumption that your first part is correct.
answered Nov 30 '18 at 4:39
Thomas ShelbyThomas Shelby
2,250220
$endgroup$
Hint: You can easily find out the coordinates of $Q_P$ by solving the system of linear equations
begin{align}
y-x=6\
x+y=8
end{align}
Knowing $Q_P$ you can express the distance between $P$ and $Q_P$ in terms of $(x_P,y_P)$.
For the third part, first find out the distance between $A$ and $B$. Let that distance be $b$. Let the distance between $P$ and $Q_P$ be $h$. Then the area of triangle is $frac{bh}{2}$. Note that the only variables are $x_P$ and $y_P$. Now use the relation $y=x^2$ and take the derivative of area with respect to either $x$ or $y$. (Depends on your substitution.) I hope you know how to proceed.
P.S:My answer is based on the assumption that your first part is correct.
answered Nov 30 '18 at 4:39
Thomas ShelbyThomas Shelby
2,250220
answered Nov 30 '18 at 4:39
Thomas ShelbyThomas Shelby
2,250220
answered Nov 30 '18 at 4:39
Thomas ShelbyThomas Shelby
2,250220
answered Nov 30 '18 at 4:39
Thomas ShelbyThomas Shelby
2,250220
2,250220
add a comment |
add a comment |
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$begingroup$
There are many points on the parabola, bellow L. From your answer to a) it is clear you have chosen one particular point. That's why your coefficient 8 is fixed and you cannot continue. The point P has to be arbitrary. In the equation of $L_p$ should be a parameter.
$endgroup$
– user376343
Nov 30 '18 at 5:18