a plant nursery sells two sizes of oak trees to landscapers. large trees cost the nursery $130 from the…

a plant nursery sells two sizes of oak trees to landscapers. large trees cost the nursery $130 from the grower. small trees cost the nursery $80. the profit for each large tree sold is $35 and the profit for each small tree sold is $26. the monthly demand is at most 450 oak trees. furthermore, the nursery does not want to allocate more than $2,000 each month on inventory for oak trees. (a) determine the number of large oak trees and the number of small oak trees that the nursery should have in its inventory each month to maximize profit. (b) what is the maximum profit? (c) if the profit on large trees were $60, and the profit on small trees remained the same, then how many of each should the nursery have to maximize profit? let x represent the number of large oak trees. let y represent the number of small oak trees. write an objective function. since profit is to be maximized, write a profit function. (a) determine the number of large oak trees and the number of small oak trees that the nursery should have in its inventory each month to maximize profit. define the relevant variables. monthly profit: z = (monthly profit to sell x large oak trees) + (monthly profit to sell y small oak trees)
Answer
Explanation:
Step1: Define variables
Let $x$ be the number of large oak trees and $y$ be the number of small oak trees.
Step2: Determine profit - per - tree values
Profit on large trees is $$60$ and on small trees is $$26$.
Step3: Write the objective function
The profit function $z$ to be maximized is $z = 60x+26y$.
Step4: Consider constraints
The nursery has a monthly demand of at most 450 oak trees, so $x + y\leq450$. Also, non - negativity constraints $x\geq0,y\geq0$. To find the maximum profit in part (b), we need more information about the constraints (such as if there are limitations on inventory cost etc.). But if we assume we only have the demand constraint $x + y\leq450$ and want to maximize $z = 60x+26y$. Since the profit per large tree is higher, we should sell as many large trees as the demand allows. So if $x = 450$ and $y=0$, the maximum profit $z=60\times450+26\times0=$27000$. For part (c), if the profit on large trees remains $$60$ and on small trees remains $$26$, and we want to maximize profit with the given demand constraint $x + y\leq450$, we still want to sell as many large trees as possible. So the nursery should have $x = 450$ large trees and $y = 0$ small trees in inventory to maximize profit.
Answer:
(a) The relevant variables are $x$ (number of large oak trees) and $y$ (number of small oak trees). The objective function is $z=60x + 26y$. (b) $$27000$ (c) 450 large trees and 0 small trees.