Jeanette R.

asked • 03/04/18# PLEASE HELP!!! question: Two positive numbers have the property that one more than the first number plus four less than three times the second, totals to 100.

(i) If we call the first number x, and the second z, write an equation that relates

these numbers.

(ii) Use (i) to write an equation that gives the second in terms of the first.

(iii) Use (ii) to write an expression for y=f (x), the product of these two numbers, in

terms of x.

(iv) Now write the product formula in vertex form, by using the completing the

square technique. (HINT: It will be much easier if you start out multiplying both

sides of (iii) by -3.)

(v) Use (iv) to graph the function in (iii). Label the coordinates of the maximum

product, as well as both x-intercepts.

(vi) Find both numbers, as well as the largest product two numbers can have, under

these conditions.

(vii) Write a complete sentence explanation of the answers to (vi)

these numbers.

(ii) Use (i) to write an equation that gives the second in terms of the first.

(iii) Use (ii) to write an expression for y=f (x), the product of these two numbers, in

terms of x.

(iv) Now write the product formula in vertex form, by using the completing the

square technique. (HINT: It will be much easier if you start out multiplying both

sides of (iii) by -3.)

(v) Use (iv) to graph the function in (iii). Label the coordinates of the maximum

product, as well as both x-intercepts.

(vi) Find both numbers, as well as the largest product two numbers can have, under

these conditions.

(vii) Write a complete sentence explanation of the answers to (vi)

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## 2 Answers By Expert Tutors

(i) x+1+(3z-4) = 100

**x + 3z = 103**

Solving for z:

(ii)

**z = (1/3)(-x + 103)**Here we just write xz then substitute (ii) for 'z':

(iii) f(x) = xz = x(1/3)(-x + 103) =

**(1/3)(-x**^{2}+ 103x)(iv)

Vertex form is g(x) = a(x-h)2 + k where (h,k) is the vertex

Re-write f(x) from (iii) as by factoring out a -1 and then add

((1/2)103)

^{2 }(½*coefficient of 'x' term): f(x) = (-1/3)(x

^{2 }- 103x + (103/2)^{2}) - (-1/3)(103/2)^{2}**f(x) = (-1/3)(x - (103/2)**

^{2}) + (1/3)(103/2)^{2}(v) The graph is a parabola pointing down, one leg passes through the origin, the vertex is at ((103/2)

^{2}, (1/3)(103/2)^{2}), and the other leg passes through the x-axis at x=103 (see below where x-intercepts are found).x-intercepts:

Setting f(x) = 0 and solving for x, we find two solutions:

(-1/3)(x-103/2)

^{2}+ (1/3)(103/2)^{2}= 0Re-arrange, multiply both sides by -3:

(x-103/2)

^{2}= (103/2)^{2}Take square root:

x-(103/2) = ±(103/2)

x = 0 and x = 103 are both solutions

We can find z for each of these x values (using (ii)):

**x=0 —> z=103/3**(check these in (i))

**x=103 —> z=0**

The largest product is the y-value at the vertex:

**(1/3)(103/2)**^{2 }or approx. 884.08(vi) The entire curve represents the possible products, xz. Since the parabola opens downward, the maximum value of xz is at the vertex.

Al P.

tutor

I assumed x and z are

*nonnegative*(zero is neither positive nor negative and I assumed they can be zero). For nonnegative x and z, the parabola is cut off where it meets the x-axis (and the graph includes the x-axis).If they must be strictly positive, then the graph does not include the x-axis and there are no x-intercepts.

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03/05/18

This question was posted earlier as well. Have a look at

https://www.wyzant.com/resources/answers/458099/two_positive_numbers_have_the_property_that_one_more_than_the_first_number_plus_four_less_than_three_times_the_second_totals_to_100

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Mark M.

03/04/18