Optimization problem.?

Question

A company produces both large and small cabinets. A small cabinet requires 1 hour of labor a day and a large cabinet requires 4 hours a day. At most 80 hours a day of labor are available. No more than 60 small cabinets and 15 large cabinets can be produced in a day. If the company's profits are $120 per small cabinet and $250 per large cabinet , how many of each should be produced to maximize profit? What is the maximum profit?

Thank you for your help.

Answer 1

I'll get you started ...

first find your constraints (hours of labor per day) and number of cabinets of each type per day.
next find your variables ( let x = number of small cabinets) and (y= number of large cabinets)
go to your calculus book and follow the examples on constrained optimization. your graph is accually going to of profit vs numbers of each type. To find your maximum profit it is going to be the same as finding an (absolute max) of your function.
I am hoping your in calc 3 as you will be able to set up this function with 3 variables and do it that way.

Answer 2

The equation of a line is y = mx + b meaning at (6, 10), the equation is 10 = m(6) + b, or 10 = 6m + b or b = 10 - 6m The x-intercept of this line represents the backside of the triangle. The x-intercept is discovered via making y = 0, so y = mx + b 0 = mx + b -b = mx x = -b/m <==== the backside. The y-intercept represents the top of the triangle, discovered via making x = 0. y = mx + b y = b <====== the top The formula of the element of the triangle is (a million/2)(base)(top). consequently, A = (a million/2)(-b/m)(b) A = (a million/2)(-b^2 / m) A = (-a million/2)(b^2/m) yet all of us know that b = 10 - 6m, so A = (-a million/2)( [10 - 6m]^2 / m ) So this may well be our section formula, A(m) A(m) = (-a million/2)( [10 - 6m]^2 / m ) we would prefer to shrink the section, so we take the by-product as commonly used. A'(m) = (-a million/2) [ 2(10 - 6m)(-6)m - (10 - 6m)^2 (a million) ] / [ m^2 ] A'(m) = (-a million/2) [ (-12m)(10 - 6m) - (10 - 6m)^2 ] / [ m^2 ] A'(m) = (-a million/2) [ (10 - 6m)(-12m - (10 - 6m)) ] / [ m^2 ] A'(m) = (-a million/2) [ (10 - 6m)(-12m - 10 + 6m) ] / [ m^2 ] A'(m) = (-a million/2) [ (10 - 6m)(-6m - 10) ] / [ m^2 ] Now, we make A'(m) = 0. 0 = (-a million/2) [ (10 - 6m)(-6m - 10) ] / [ m^2 ] extreme values: 5/3, -5/3 we are able to reject 5/3 because of the fact it fairly is a favorable slope (the only thank you to style triangles from the 1st quadrant is that if the slope is detrimental). That leaves -5/3, so this tells us the minimum section happens while the slope m = -5/3. to locate the equation of the line, all we would desire to consistently do is locate severe college undemanding approaches to locate the equation of the line with slope m = = -5/3 and for the time of (6, 10). (y2 - y1)/(x2 - x1) = -5/3 (y - 10)/(x - 6) = -5/3 y - 10 = (-5/3)(x - 6) y - 10 = (-5/3)x + 10 y = (-5/3)x + 20