Both of these seem easy, but I can't seem to figure them out?

All of this is Grade 12 Calculus (Derivatives)





1. If the sum of two positive numbers is 8, show that the square of one number added to the cube of the other is at least 44.





2. Dianna wants to enclose a rectangular flower pot that adjoins a brick retaining wall. If she only has 30 m of decorative fencing, what should the dimensions of the flower pot be so that she has a maximum area?

Both of these seem easy, but I can%26#039;t seem to figure them out?
1)


x+y = 8


x = 8 - y





x^2 + y^3 = (8-y)^2 + y^3


= y^3 + y^2 - 16y + 64


Let%26#039;s call that function f(y)





We are required to find the minimum value of f(y).


f%26#039;(y) = 3y^2 + 2y - 16 = (3y + 8)*(y - 2)


Since y %26gt; 0, this function is minimum when y = 2


f(2) = 44


so 44 is the minimum value of that above problem.





2) Without proof, the second part will have max area when it%26#039;s a square.


ie 7.5 x 7.5 sides.
Reply:If one of the sides is the retaining wall, the answer to Q:2 is wrong! Report It

Reply:1 The equation is x^3 + (8-x)^2.


This becomes: x^3 + X^2 -16x +64.


Derivative is 3x^2 + 2x - 16.


Solving the quadratic formula you get to (-2 +_ 14)/6.


Solving for positive numbers, the answer is 12/6, or 2.


That%26#039;s your local maximum. 2^3 + (8 -2)^2 = 8 + 36 = 44.





2 The largest rectangle in area for a perimeter of x will always be a square of x/4 on a side. Since you want to maximize with material for three sides available, the largest area will be to form a square of 10m per side.
Reply:Amswer to Question 1:


If (x+y)=8, x%26gt;=1, y%26gt;=1, and x and y are integers


Show (x^2+y^3)%26gt;=44





Answer: x=8-y


So


x^2+y^3=(8-y)^2+y^3


=64+y^2-16y+y^3


So when is y^2-16*y+y^3 minimum - at y=2


Why


d/dy(y^2-16*y+y^3)=0 gives 2*y-16+3*y^2=0 gives y=2, -8/3


Only y=2 is acceptable as y is postive.


d/dy(2*y-16+3*y^2)=2+6*y%26gt;0 at y=2, hence a minimum





So x^2+y^3%26gt;=64+2^2-16*2+2^3=44








Answer to Question 2:


If x is the length along the retaining wall and y is length perpendicular to it.





x+2*y=30


Maximize xy





Maximize (30-2*y)*y=30*y-2*y^2





d/dy(30*y-2*y^2)=30-4*y. Put this equal to zero gives


30-4*y=0 implies y=30/4=7.5





So y=7.5 and x=30-2*y=30-2*7.5=15





To show that it maximizes d2/dy2(30*y-2*y^2)=-4 which is less than zero. So the value of y=7.5 maximizes the area.





ASK QUESTIONS SEPARATELY. YOU WILL HAVE A BETTER CHANCE OF GETTING THE BEST ANSWER!
Reply:Hmmmm! Ask somebody else.
Reply:x+y= 8


let A = x^2 + y^3 = x^2 + ( 8 - x)^3


dA/dx= 2x - 3(8 - x )^2 =2x - 192 + 48 x -3 x^2= -3x^2 + 50x -192


2nd derivative = -6x + 50


let dA/dx = 0


3x^2 - 50x + 192 = 0


(3x - 32 )( x - 6) = 0 then x = 6 and y = 2


at x = 6 the 2nd derivative= positive number


at x = 6 , A = min. value


min. of A = x^2 + y^3= 6^2 + 2^3 = 44



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