MATH SOLVE

5 months ago

Q:
# If Florist B increases the cost per rose to $5.20,for what number of roses is it less expensive to order from Florist A? From Florist B?

Accepted Solution

A:

Answer:For 38 roses at $5.15 per rose, Florist A is Less Expensive then Florist B For 34 roses at $5.20 per rose, Florist A is Less Expensive then Florist B Step-by-step explanation:The Full Question Reads:Derek wants to order some roses online. Florist A charges $4.75 per blue rose plus $40 delivery charge. Florist B charges $5.15 per red rose plus $25 delivery charge. If Florist B increases the cost per rose to $5.20, for what number of roses is it less expensive to order from Florist A? From Florist B?To begin we need to understand our given information and what we are actually looking for. Given InformationFlorist A:charges $4.75 per blue rosecharges $40 per deliveryFlorist B:charges $5.15 per red rosecharges $25 per deliveryNote: in this problem we ignore the colour of the rose (red or blue) as it does not affect our solution or contributes to it. Therefore we shall call our first variable representing the number of roses as [tex]x[/tex]. The next step is to construct a system of equations representing our problem and our given information. Here we are looking at linear relationships so a linear function of the form:[tex]y=ax+b[/tex] Eqn(1).will suffice where[tex]y[/tex]: represents the total cost from the Florists (i.e. number of roses and delivery charges). Dependent Variable[tex]x[/tex]: represents the number of roses purchased from the Florists. Independent Variable[tex]a[/tex]: is our relationship factor between [tex]y[/tex] and [tex]x[/tex][tex]b[/tex]: is our constant which in this problem is the delivery charge value for each florist. Since we have two Florists, we will construct two equations, one for each florist, by employing Eqn(1). and our given information as follow:Florist A: [tex]y=4.75x+40[/tex] Eqn(2).Florist B: [tex]y=5.15x+25[/tex] Eqn(3).By employing both equations above and writing them as an inequality since we are looking for which value of [tex]x[/tex] (i.e. number of roses) will the less expensive Florist be and then solving for [tex]x[/tex] we have: [tex]4.75x+40<5.15x+25\\[/tex][tex]4.75x-5.15x<25-40\\[/tex] Gathering all similar terms together[tex]-0.4x<-15\\[/tex] Simplifying[tex]x>\frac{-15}{-0.4}\\[/tex] Solving for x [tex]x>37.5[/tex](note how < changes to > since any multiplication/devision process of negative sign in an Inequality will change the order of < to > and vice versa). Since a rose has to be sold as a whole and not half we will say that [tex]x>38[/tex] So we can then plug in the value of [tex]x[/tex] in Eqn(2) and Eqn(3) and find the cost of buying more than 38 roses from each:Eqn(2): Florist A: [tex]y(x>38)=4.75*38+40=220.5[/tex] Eqn(3): Florist B: [tex]y(x>38)=5.15*38+25=220.7[/tex]Which tells us that Florist A is less expensive than Florist B by $0.20 for a purchase of more than 38 roses. Next the question tells us that Florist B increases the cost from $5.15 to $5.20 per rose, which in Eqn(3) denotes our [tex]a[/tex] value and thus Eqn(3). now becomes:Florist B: [tex]y=5.20x+25[/tex] Eqn(3).Applying the same method like before and solving for the value of [tex]x[/tex] we have: [tex]4.75x+40<5.20x+25\\[/tex][tex]4.75x-5.20x<25-40\\[/tex] Gathering all similar terms together[tex]-0.45x<-15\\[/tex] Simplifying[tex]x>\frac{-15}{-0.45}\\[/tex] Solving for x [tex]x>33.3[/tex]Similarly as a rose has to be sold as a whole and not half we will say that [tex]x>34[/tex] So we can then plug in the value of [tex]x[/tex] in Eqn(2) and Eqn(3) and find the cost of buying more than 34 roses from each:Eqn(2): Florist A: [tex]y(x>34)=4.75*34+40=201.5[/tex] Eqn(3): Florist B: [tex]y(x>34)=5.15*34+25=201.8[/tex]Which tells us that Florist A is again less expensive than Florist B by $0.30 for a purchase of more than 34 roses. It was also shown that as the cost of the rose increased by $0.05 the number of roses for purchase decreased.