Dhelia's garden

Let $R(n), O(n),$ and $J(n)$ denote the amount of space in square meters required to plant $n$ Rose, Orchid, or Jasmine trees, respectively. We will now find the increase of each function when adding one more tree.

$O(n + 1) = 5(n + 1) = O(n) + 5$

$R(n + 1) = (n + 1)^2 = R(n) + 2n + 1$

$J(n + 1) = (n + 1)^3 = J(n) + 3n^2 + 3n + 1$

For minimal garden space, we will increase the most efficient functions at each stage, until we reach $10$ trees. We will compare the efficiency of the other trees to the Orchids, because $O(n)$ has a constant increase of $5$ (square meters) per tree.

When is the increase in $R(n)$ less than of $O(n)$ ?

$2n + 1 < 5 \Rightarrow n < 2 \:$ , meaning that we can take two Rose trees.

When is the increase in $J(n)$ less than of $O(n)$ ?

$3n^2 + 3n + 1 < 5 \Rightarrow n < 1 \:$ , meaning that we can take one Jasmine tree.

We now have a total of three trees, and the rest will be seven Orchid trees. Summing the areas, we get,

$R(2) + J(1) + O(7) = 4 + 1 + 35 = \boxed{40}$

Clearly we don't want to have too many jasmine trees, as that would take up a lot of space.

Let's see what we can do with no jasmine trees.

We want to minimize $a^2+5b$ where $a$ and $b$ are integers that sum up to 10. If we have $a=2$ or $a=3$ , this will be minimized at a value of 44.

Now let's have one jasmine tree. This is 1 square meter.

Now we want to minimize $a^2+5b$ , this time where they sum up to nine. This is minimized at $a=2$ . We have $2^2+7\times5+1^3=40$ .

Now let's try two jasmine trees. This is 8 square meters.

This time $a$ and $b$ sum to 8, minimized at $a=2$ and $b=6$ , giving $2^2+6\times5+8=42$ . Uh-oh, we're getting bigger.

Clearly getting even more jasmine trees is not the solution. The minimum amount of space is therefore equal to $\boxed{40}\ \blacksquare$ .

Let $a, b, c$ be the number of rose orchid and jasmine trees.As we have one equation given but three variables. The area required is p $a+b+c=10$ $a^{2}+5b+c^{3}=p$ $p=50+(a^{2}-5a)+(c^{3}-5c)$ Deciding the values for a and c for the minimum value of $**(a^2-5)** and **c^3-5c*$ * respectively independently. Take into consideration that a and c are non-negative integers we get the value of p as $40$

Treat the problem like a marginal problem in economics. Find the lowest additional area required for each sucessive tree. The first one will be either a jasmine or rose, each requiring 1 square meter, and the second would be the other one. Third would be a rose tree, because going from 1^2 to 2^2 is 3, which is less than 5 or 7 (for orchid or jasmine). After the 2 rose and one jasmine, the rest must be orchid, because that has the lowest value. So 7 orchid. 7 * 5 + 2^2 + 1 = 35 + 4 + 1 = 40.

Okay so I thought it like this: Consider the equation: a+b+c = 10, a, b, c are non-negative integers.

Now, we need to minimize $a^{2}$ + 5b + $c^{3}$

Clearly, we have the inequality c <= a <= b. Setting c = 1 and trying out the few initial cases, answer comes out to be 40.

Firstly- THIS QUESTION IS NOT COMBINATORICS- IT'S CALCULUS

Solution: $a + b + c=10$

$b =10-{a-c}$

let $D$ denote the Area

$D= a^2+5b+c^3$

Subbing in rearranged expression for $b$ we get

$D= a^2+5(10-a-c)+c^3$

$D= a^2-5a+c^3-5c+50$

For now as long as $a+b \leq 9$ we're fine- we can think of $a$ and $c$ as being independent.

It is just a case of differentiating which I won't bother to type and finding the nearest integer values to where these these minimum points occur.

So minimum for $a^2-5a$ occurs at $a= \frac{5}{2}$ nearest value is $2$ (or $3$ gives the same answer) Minimum is $2^2-5(2)=-6=3^2-5(2)$

Minimum for $c^3-5c$ occurs at $c^2=\frac{5}{3}$ after differentiating nearest integer $c$ is $1$ or $2$ .

$1^3-5(1)=-4$ and $2^3-5(2)=-2$ so minimum is $-4$ .

So minimum for $D$ is $-6-4+50=\boxed{40}$

We can introduce 3 variables r,o,j which represent the number of each type of tree. This allows us to form an equation for the area A which we can then minimise. $A = 5o +r^{2} +j^{3}$ We have the constraint such that $r + o + j = 10$ We can then rearrange for o $o = 10 - r - j$ And substitute this into the original equation $A = j^{3} + r^{2} -5j -5r + 50$ $A = j(j^2-5) + r(r-5) + 50$ Now we minimise A for this we know that j and r have to be greater than 0 By inspection the minimum returned from $j(j^2-5)$ Is $-4\hspace{2 mm} when \hspace{2 mm} j=1$ The minimum returned from $r(r-5)$ Is $-6\hspace{2 mm} when \hspace{2 mm} r=2 \hspace{2 mm} or \hspace{2 mm} r=3$ We know that the sum has to be 10 $r + o + j =2+ o + 1 = o + 3$ or $r+o+j=3 + o + 1= o + 4$ $o = 7 \hspace{2 mm} or\hspace{2 mm} o = 6$ Subbing these values into A $A = 5(7)+ 2^2 + 1^3 = 40$ and $A= 5(6) + 3^2 + 1^3 = 40$ Hence minimum A is 40

Let $a,b$ and $c$ be the numbers of rose, orchid and jasmine trees we require, accordingly.

Note that since the area required for rose trees and jasmine trees grows exponentially, they remain optimal up until a certain number. In this case, that number is $5$ , since the area required for the orchid trees increased by $5$ for each tree, meaning that a single tree will need $5$ square meters. Every area smaller than that is in our favor.

Therefore, we may take rose and jasmine trees until the area required for them goes above $5$ .

Rose trees

• For $1$ , we require $1$ square meter, which is less than $5$ .
• For $2$ , we require $4$ square meters, which is less than $5$ .
• For $3$ , we require $8$ square meters, which is greater than $5$ .

We are have to take exactly $2$ rose trees in order for their area to be smaller than $5$ , meaning that $a=2$ .

Jasmine trees

• For $1$ , we require $1$ square meter, which is less than $5$ .
• For $2$ , we require $8$ square meters, which is greater than $5$ .

We may take a single jasmine tree in order for the area required for it to be smaller than $5$ , meaning that $c=1$ .

Orchid trees

The rest of the trees we require will be orchid, since any higher value for the previous two will result in greater area. Thus, $b=10-3 \Rightarrow b=7$ .

Finally, the total area for this set of trees would be $P=a^{2}+5b+c^{3} \Rightarrow P=40$ .

the min. values of a^2 and c^3 would be 1 so let us assume 1 rose and 1 jasmine and hence 10-2=8 orchid trees. so total area req.=1^3 + 1^2 +5 8 =42. let us assume 2 rose trees 1 jasmine tree now orchid will be 7 hence total area req. 2^2=4 + 1^3=1 + 7 5=35 which sum upto 40. which is min. area

well.... the most rapidly increasing quantity is c^3 coz its derivative being 3c^2 so increasing no. of jasmines will never give me a mininum.... so taking c=1 now rate of change of a^2 being 2a .... i'd like to keep 2a<5b so i assume two values of a i.e 2 &3

for both these values the result of a^2 + 5b + c^3 = 40

3 apple trees + 1 jasmine tree + 6 orchid trees

I'll just say that if you already built n trees and going to build 1 more trees, how many more sq.m. do you need.

Here's the pattern of number of sq.m. do you need more to build. (Start from building 1 tree, then how many more.)

Rose: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. (Why is it increasing like that? $1+3+5+...+(2n-1) = n^{2}$ .)

Orchid: 5, 5, 5, 5, 5, 5, 5, 5, 5, 5.

Jasmine: 1, 7, 19, 37, 61, 91, 127, 169, 217, 271.

You can choose to build any trees, but we don't need to increase the number of spaces we need.

In this one, you choose rose first, then choose jasmine, then choose rose, and so on....

Start(1, 5, 1) -> (3, 5, 1) -> (3, 5, 7) -> (5, 5, 7)

Do this 10 steps and we get {R,J,R,R,O,O,O,O,O,O} or {R,J,R,O,O,O,O,O,O,O}.

Therefore, the minimum $= 3^{2} + 5\times 6 + 1^{3} = \boxed{40}$ . (another one also gets 40.)

grow 2 rose trees, 7 orchid and 1jasmine tree....because cube of any no. produces large no. so we must take cube of less no.as we have taken only one...

sqrt = raiz quadrada

A = x² + 5y + z³ x + y + z = 10 --> y = 10 - x - z

portanto A = x² - 5x + z³ - 5z + 50

dA/dx = 2x - 5 = 0 --> x = 2,5 dA/dz = 3z² - 5 = 0 --> z = sqrt(5/3)

P = (2,5 ; sqrt(5/3))

d²A/dx² = Axx = 2 d²A/(dx . dz) = Axz = 0 d²A/(dz . dx) = Azx = 0 d²A/dz² = Azz = 6z

H(2,5 ; sqrt(5/3)) = Axx(2,5 ; sqrt(5/3)) . Azz(2,5 ; sqrt(5/3)) - Axz(2,5 ; sqrt(5/3)) . Azx(2,5 ; sqrt(5/3)) H(2,5 ; sqrt(5/3)) = 2 . 6 . sqrt(5/3) H(2,5 ; sqrt(5/3)) = 12 . sqrt(5/3) > 0

Axx(2,5 ; sqrt(5/3)) + Azz(2,5 ; sqrt(5/3)) = 2 + 6 . sqrt(5/3) > 0

H > 0 e Axx + Azz > 0 então P = mínimo

substituindo o ponto P na área A(2,5 ; sqrt(5/3)) = 39,45

arredondando A(2,5 ; sqrt(5/3)) = 40

Please, advise me if im wrong. The question can be re-written as follows:

"Find the minimum value of $a^{3} + b^{2} + 5c$ where $a + b + c = 10$ and they are non-negative integers".

$5n < n^{2}$ if $n ≥ 6$ . So, if I want to minimize, shouldn't I take $c = 6$ ? (this is the point in which I want you to advise me)

Then, our expression turns to be $a^{3} + b^{2} + 30$ , with $a + b = 4$ . We can check the five cases, but clearly the minimum is when $a = 1$ and $b = 3$ . This yields $1^{3} + 3^{2} + 30 = {\color{#20A900}{\boxed{40}}}$ .

First, you want to keep the Jasmine Trees as Small As Possible since they cover the most space. Keep it at 1. Since There are 9 trees left, you want to use a 3 - 6 split. Since 3^2 is less than 6^2, use 3^2 instead. That gives you 9 so plant 3 rose trees. 9+1=10. The Remaining 6 "seeds" can be used to plant orchid trees which requires a decent amount of space. 5(6)= 30. 9 + 30 + 1= 40.